Which ones are left out?

If $n=2k+1$ we can take the triplet $(2k+1, 2k^2+2k, 2k^2+2k+1)$ , of course this triplet works for $k\geq1$

If $n=2k$ we can take the triplet $(2k, k^2-1, k^2+1)$ and this triplet works for $k\geq 2$

It only remain to prove that 1 and 2 do not appear as one of the entries in a Pythagorean triple.

If $1^2+b^2=c^2$ then $(c+b)(c-b)=1$ , thus $b=0$ which is not possible.

If 2 appears in the Pythagorean triple $(a,b,c)$ then $c\neq 2$ and if $a=2$ we get $2+b^2=c^2$ and then $(c+b)(c-b)=2$ which is not possible since $c+b$ and $c-b$ have the same parity.

If n is odd and >1, then $n^2 + \left(\frac{n-1} 2\right)^2 = \left(\frac{n+1} 2\right)^2$ so n is part of a Pythagorean triplet.

If n=2k, and k is part of a Pythagorean triplet, then so is n=2k since if (a, b, c) is a Pythagorean triplet, then so is (2a, 2b, 2c).

Finally, 4 is part of a Pythagorean triplet since $3^2 + 4^2 = 5^2$ .

Combining the above facts, the only remaining numbers are 1, 2.

For any even number the triplet is = 2n , n^2-1 , n^2 +1 For odd number = 2n+1 , 2n(n+1) , 2n(n+1) +1 For 1,2 we get one of the sides as zero ...rest all numbers are possible Hence sum=1+2=3

Assume $a$ as the smallest number in the Pythagorean triple. If $a$ is an odd number, then $a^2$ is an odd number, too. Because of that, we can write $a^2$ as $2k+1$ with $k$ is a positive integer. Now, we use the fact that $(k+1)^2 - k^2 = 2k+1.$ If $a^2 = 2k+1$ , then $a^2 + k^2 = (k+1)^2$ which make it a Pythagorean triple. If $a$ is an even number, remember that an even number is the multiple of an odd number. We can use the fact that, if $a,b,c$ is a Pythagorean triple, then $xa, xb, xc$ is a Pythagorean triple, too if $x$ is a non-zero integer. For the odd number, the only positive integer that cannot appear as one of the entries in a Pythagorean triple is $1$ , because $1 = 2*0 + 1$ , thus make $k = 0$ while $k$ must be a positive integer to make the number appear as one of the entries in a Pythagorean triple. Because $1$ cannot appear as one of the entries in a Pythagorean triple, $2$ cannot appear as one of the entries, too. After all, $2$ is the multiple of $1$ . Then why the power of $2$ , like $4, 8$ , etc can be appear as one of the entries in a Pythagorean triple? That's because $4$ is one of the entries in Pythagorean triple $3,4,5$ . Because of that, the power of $2$ , except the $2$ itself can be appear as one of the entries in a Pythagorean triple. So, there are only two positive integers which do not appear as one of the entries in a Pythagorean triple. They are $1$ and $2$ . So, the sum of those integers are $3$ .

Since all pairwise triples of integers satisfying $x^2 + y^2 = z^2$ are given by $x = |u^2 - v^2|, y = 2uv$ , and $z = |u^2 + v^2|$ with $gcd(u,v) = 1, u\neq v mod 2$

Every integer can be represented by even or odd parity so there will be two cases.

$Case 1. \forall u,v \in Z^+$ we will have an even $2uv$ in the equation except for $2$ , as $p>q$ .

$Case 2. \forall u,v \in Z^+$ we will have $|u^2 - v^2| = (u-v)(u+v)$ . Suppose that $1$ is included in the equation, so $(u-v) = (u+v) \Rightarrow 2u = 0 \Rightarrow u = 0$ , a non-positive integer - Contradiction!

Hence, we get sum of the leftover integers $1+2 = 3$ .

a=3, b=4, c=5 is the most basic Pythagorean triple. And it also works for 4,6,8 and 5,10,15 and n, 2n, 3n... Which the range of n is integer from 3 to infinity...

So, the positive integers that fulfill the criteria are only 1 and 2 = 3

We will take the sides of the right-angled triangle to be $a$ = $m^{2}-n^{2}$ , $b$ = 2 $mn$ and $c$ = $m^{2}+n^{2}$ . Obviously $m^{2}+n^{2}$ is the greatest side.

Let us assume $m^{2}-n^{2}$ as the smallest side. Since we know the least possible non-zero(since side cannot be zero) difference two squares occurs when they are consecutive and both smallest and equals 3(= $2^{2}-1^{2}$ ), so we can obviously rule out 1 and 2 as sides of the triangle.

Now we assume $m$ and $n$ are both odd( This is equivalent to taking both $m$ and $n$ even). So $m+n$ and $m-n$ are both even and so is $m^{2}+n^{2}$ . That implies $a$ = $m^{2}-n^{2}$ = even, $b$ = 2 $mn$ is always even and $c$ = $m^{2}+n^{2}$ is also even. That implies every even positive integer except 2 is always a part of a Pythagorean triple.

Next we take $m$ as odd and $n$ as even( Equivalent to $m$ as even and $n$ as odd). So $a$ and $c$ are odd, as always $b$ is even. So all the odd positive integers will also be a part of a Pythagorean triple.

Using the above results we find that the only positive integers not entering in a triple are 1 and 2 so their sum = $1+2$ = $3$

If (a,b,c) form a Pythagorean triplet, it is evident that for an integer n, (na, nb, nc) will also form Pythagorean triplet.

Let, a^2 + b^2 = c^2.

This gives a^2 = (c+b)(c-b). Now, if a is a multiple of 2, then there are 2 possibilities: a = 2 or a = 2m where m is a natural number. Clearly, a=2 will never form a Pythagorean triple because if it formed so, then it would imply c+b=4, c-b=1 as both c+b and c-b can't be equal to 2. This gives non-negative solutions for c and b and hence 2 doesn't satisfy.

Now, let's come to 2m. If m is even then a = 4p which is a multiple of 4 and as (3,4,5) is a triplet, a will form a trip-let also. If m is odd then a will not belong to a triplet if we can show that m also doesn't belong to any triplet. Again, it is best to start with prime m, as if m is composite, then again we need to prove that the prime factors of m will never form a Pythagorean triplet.

However, for all prime m's, m^2 = x^2 - y^2 will give integer solutions of x and y as x+y = m^2 which is odd and x-y = 1 which is also odd. Thus all primes except 2 will get into Pythagorean triplet. Their multiples also will form Pythagorean triplet.

Now, let's come to 1. Only possibility is (1,0,1) or its combinations but 0 is not natural number. Hence, 1 is also not forming Pythagorean triplet.

So sum = 1+2 = 3

This is a test. You should submit your solutions :)

Every integer greater than two is part of a primitive or non-primitive Pythagorean triple. Therefore, 1 and 2 are the only integers that are not part of any Pythagorean triple and 1 + 2 = 3.

Any Pythagorean triple is of the from $m^2-n^2, 2mn , m^2+n^2$ It is easy to see that all odd and even numbers are a part of some or the other pythagorean triple with the exception of $1,2$ where one of the terms in the Pythagorean triple becomes $0$ .

By the Euclides formula we can assume that , a= m^{2} - n^{2} , b = 2mn, c = m^{2} + n^{2}, where m and n are natural numbers ,and m>n.

Therefore n >=1 and m>1, but b is a multiple of 2, so all even numbers that satisfies the conditions can be entered in the triple .

Furthermore a =m^{2} - n^{2}, and we can show that all odd numbers can be writted when m= n+1, so a= (n+1)^{2} - n^{2}, a= 2n +1. The pairs (m,n) that don`t satisfies the conditions are: (1,0) , (0,0) , (1,1). For that we conclude that a or b are different of 1 ,2 and 0, and the sum is 3

Every integer greater than two is part of a primitive or non-primitive Pythagorean triple. Therefore, 1 and 2 are the only integers that are not part of any Pythagorean triple and 1 + 2 = 3.

From http://en.wikipedia.org/wiki/Pythagorean_triple
Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple.

Therefore, the only integers that are not part of a Pythagorean triple are 1 and 2

Sum = 3

Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple.

Therefore, the only integers that are not part of a Pythagorean triple are 1 and 2 So, Sum = 3

All Pythagorean triple can be written as $(2mn,m^2-n^2,m^2+n^2)$ for some integer $m>n>0$
Proof: $(2mn)^2 + (m^2-n^2)^2 = m^4 + 2m^2n^2 + n^4 = (m^2+n^2)^2$
For the minimum value of m $(2)$ and n $(1)$ : the triple is $(4,3,5)$
Any even number $x > 2$ can be written as $2.m.1$ cause we can easily find an $m > 1 > 0$
Any odd number $y > 1$ can be written as $m^2-(m-1)^2=(m-(m-1)).(m+(m-1))=2m-1>1$ thus $m>1$ so $m>m-1 \geq 1 > 0$ satisfying the conditions above.
Obviously, only $1$ and $2$ can't be one of the entries in any Pythagorean triple.

Para se calcular os ternos pitagóricos, utilizamos a hipotenusa como /m^2+n^2/ e os catetos como /2 m n/ e /m^2-n^2/.

Usando-se n=1 podemos encontrar quaisquer números por uma das três fórmulas, exceto o 2, pois implicaria em ter m=n=1 e encontraríamos um lado igual à zero, e, exceto o um, pois, da mesma forma, um dos lados seria zero.

Given any positive integers $m, n$ with $m>n$ we can form pythagorean triple $(a,b,c)$ with $a=m^2-n^2, b=2mn, c=m^2+n^2$ . So choosing $m+n=p$ and $m-n=1$ we can make a triple with any prime except for 2. Multiples of a triple are also a triple. Thus the only excluded positive integers are 1 and 2

Clearly, all odds (except 1) are part of a Pythagorean triple because. If k is odd, then the Pythagorean triple is k, floor of k^2/2, ceiling of k^2/2. Multiply each of these by 2 and all 2 (mod 4) terms will be part of Pythagorean triples with the exception of 2. With 3, 4, 5, we multiple by n so that all terms divisible by 4 are part of Pythagorean triples. This means that the only 2 positive integers not part of a Pythagorean triple are 1 and 2, and 1+2=3.

Euclid’s Formula says that for the integers a,b,c to satisfy a^2+b^2=c^2 , a,b,c must also satisfy a=m^2-n^2 , b=2mn, and c=(m+n)^2.

First, look at the condition a=m^2-n^2. Any number squared can be written as sum of consecutive odd integers. Therefore, any odd integer can be created simply by subtracting two consecutive squared numbers, because everything will cancel out except the desired odd number. But, 1 cannot be created because it is the base case.

Next, looking at the condition b=2mn , we see that each Pythagorean triple has at least one even number. Even numbers can be rewritten as (2 X odd) or (2 X even). If it is in the former case, then the number is just a multiple of an odd number, which we know is included, and hence any even number of this case except 2(2=2 X 1) in that case is included.

If it’s the latter case, the even number can be rewritten as (4X(even or odd)). We know that 3^2+4^2=5^2 is a Pythagorean triple, and so any multiple of 4 will be included. Therefore, all even numbers in this case are included. So, the only numbers that are not included are 1 and 2, and their sum is 3.

Let m^2 + n^2 = (m+1) ^2, => 2m+1 = n^2 => m = (n^2 - 1) / 2

So for any positive odd integer n > 1, m is a positive integer and they form a pythagorean triple. Only when n = 1, m = 0 that doesn't form pythagorean triple.

Again Let m^2 + n^2 = (m+2) ^2, => 4m+4 = n^2 => m = (n^2 - 4) / 4

So here for any positive even integer n > 2, m is a positive integer and they form a pythagorean triple. Only when n = 2, m = 0 that doesn't form pythagorean triple.

So only 1 and 2 can't take a part of pythagorean triple. So ans is 2+1=3.

As to find a pythagorean triplet of any number ther is formula, for any odd number n , (n^2-1)/2 and (n^2+1)/2 are the other Two numbers of the pythagorean triplet of that odd number. and for even number n, (n/2)^2-1 and (n/2)^2+1 are the other two numbers of the pythagorean triplet of that even number. NOW, by applying the formula on 1 and 2 we did not get any other number of the triplet , we get only 0 and 0.Thus the sum is 1+2=3

Systematically:

First we eliminate all odd numbers

Then we eliminate all even numbers which have an odd factor,

Now, we are left only with powers of 2 But since we know that '4' has triplet , so all multiples of 4 can be made as triplets !! So, the only remaining numbers are 1 and 2 , so 1+2= 3 :)

First, we get odd number 2k+1 (k is an integer >0) and try to put it in a pythagorean triple. If we square it, we'll get 4 $k^2$ + 4k +1. And we can express 4 $k^{2}$ + 4k +1 as the difference of (2 $k^{2}$ +2k+1)^{2} and (2 $k^{2}$ +2k)^{2}. So $(2k+1)^{2}$ + $(2k^{2}$ +2k)^{2} = $(2k^{2}$ + 2k + 1)^{2}. Furthermore, we know that this is a possible triangle since it satisfies the triangle inequality. The hypothenuse is 2 $k^{2}$ +2k+1 which is less than sum of sides (2 $k^{2}$ +2k) + (2k+1) since k is always positive.

We also know that if one of a number's factors is in a pythagorean triple, then that number must also be in a pythagorean triple.

And since we've proven that all odd numbers can be in a pythagorean triple, and it is common knowledge that the even number 4 can also be a pythagorean triple, then all the multiples of these numbers must also be in a pythagorean triple. (Note that any number greater than 2 will have always have an odd factor aside from 1 (or have 4 as a factor). So we can conclude that all integers greater than 2 can be in a pythagorean triple.

That leaves us with two numbers that can never be in a pythagorean triple, 1 and 2. So adding them up, we get 3.

d d=a a+s s case 1: let a=odd => a a=odd=f.

(f-1)/2=s&&(s+1)=d

              but a,s,d are  natural nos thus a not=1

case 2: let a+even
=> a*a=even=g

                           (g/4)-1=s&&(g/4)+1=d

                              but a,s,d are  natural nos thus a not=2

                           hence nos are 1 &2

                     =>sum=1+2=3

For any odd number n except 1, we can always find numbers (n^2-1)/2 and (n^2+1)/2 which can form a pythagorean triplet with n.Thorem-each group of three numbers proportional to a pythagorean triplet always forms a pythagorean triplet.since 4 forms a part of a pythagorean triplet 3,4,5.hence ,all multiples of 4 will be a part of some pythagorean triplet.so,the remaining possibilities are odd multiples of 2 except 2 itself.But every odd multiple of 2 is also a multiple of any odd number.so,they will also be a part of some pythagorean triplet.So,the only remaining possiilities are 1 and 2 and their sum=3.

First, note that all odd positive integers $a>1$ will be part of a Pythagorean Triple of the form $(a,\frac{a^2-1}{2},\frac{a^2+1}{2})$ . (1 does not work, because it generates the non-triple (1,0,1).) Then, note that we can always scale a triple by multiplying each of its elements by 2. By these first two facts, we can tell that any number $b=2^ka$ , where $k$ is a non-negative integer, and $a$ is an odd integer greater than 1, is a part of some Pythagorean triple. Using the Unique Prime Factorization of Integers, the only numbers not of this form are powers of 2; that is, $1, 2, 4, 8, ...$ . However, 4 is part of the Pythagorean triple (3,4,5), and that can be scaled (by multiplying by 2) to any greater power of 2. Therefore, all numbers except 1 and 2 are part of some Pythagorean triple. You can also conclude that 1 and 2 are definitely NOT part of a Pythagorean triple by noting several points: a) Checking (1,1,1),(1,1,2),(1,2,2),(2,2,2) can verify that, if either number were part of a triple $a,b,c), a<=b<=c$ , then they would have to be the smallest one ( $a$ ). b) Then rearrange the Pythagorean Theorem to get $a^2=c^2-b^2=(c+b)(c-b)$ . If $c=b$ , then that would imply that $a^2=0$ , which would not work. So $c>b$ and $(c-b)>0$ . If we wanted $a=1$ , then we would need $c+b=c-b=1$ , which is not possible. If $a=2$ , we would need either [ $c+b=1,c-b=4$ ] or [ $c+b=2,c-b=2-$ ] or [ $c+b=4,c-b=1$ ], none of which are possible for $b,c$ positive integers. Therefore, 1 and 2 are the only positive integers that are not part of some Pythagorean Triple. The final answer is 1+2=3.

Every odd positive integer can be expressed as $m^{2}-n^{2}$ and every even positive integer can be expressed as $2mn$ . However, the odd number $1=1^{2}-0^{2}$ makes the other summand $0$ , and the even number $2=2\times1\times1$ makes the other summand $0$ . Since they must be strictly positive, 1 and 2 are the only positive integers that cannot be apart of a Pythagorean triple. $1+2=3$ .

Any primitive Pythagorean Triplet is of the form $(m^2 - n^2),(m^2 + n^2),(2mn)$ for positive integers m and n. Clearly, 1 and 2 cannot be put into that format.

Now, to express a number greater than 2 as a part of a pythagorean triplet: The number is either a prime or a multiple of a prime.

Case I: The number is a prime other than 2 or the multiple of a prime other than 2. We know that primes congruent to 1 modulo 4 can be expressed as $m^2 + n^2$ whereas primes congruent to 3 modulo 4 can be expressed as $m^2 - n^2$ . Thus, all primes greater than two are a part of some pythagorean triplet. Multiples of such primes can be expressed in the form $k(m^2 - n^2)$ or $k(m^2 + n^2)$ where k is some integer and $m^2+n^2$ and $m^2-n^2$ is the concerned prime. Thus, multiples of such primes are also a part of a pythagorean triplet(not necessarily primitive).

Case II: The number is a multiple of 2 alone. Since the number is greater than $2^1$ it must be divisible by $2^2$ . In other words it must be either 4 or a multiple of 4. If the number is 4 then it is a part of the triplet ( 3, 4, 5 ). If the number is greater than 4, it can be expressed as $2^x(4)$ . Thus, it can be expressed as a part of the triplet $(2^x3,2^x4,2^x5)$ .

Thus, any integer greater than 1 and 2 can be expressed as the part of a pythagorean triple. Thus, 1 and 2 are the only natural numbers that cannot be be expressed as the part of a pythagorean triple.

All odd numbers (apart from 1) are part of a Pythagorean triple. One can see this by seeing that if you have n, n^2/2+1/2, and n^2/2-1/2, it will always be a Pythagorean triple, provided that all of the above are integers, and the will all be integers if n is odd. So if I wanted to find a Pythagorean triple for 11 for example, I would just do 11, (121/2)-1/2=60, and (121/2)+1/2=61. Also, any even number can be written as 2^q * r where q is a positive integer and r is an odd number. So if I want to find a Pythagorean triple for an even number, I just find the triple for r and multiply all the numbers in the triple by 2^q. This is all well and good, but it leaves one problem- what about numbers that are just powers of two? The 3, 4 and 5 triple provides triples for powers of two above 4, but that still leaves 1 and 2, which are the only positive integers which do not fit into a Pythagorean triple.

The only integers which don't appear are 1 and 2. 3, 4, and 5 appear as part of the smallest primitive pythagorean triple. From here on, all even numbers are part of a doubled primitive triple, whereas all odds form the shortest legs of new primitive triples (the difference of two consecutive square numbers is always odd).

Thus the answer is 1 + 2, or 3.

Only 1 & 2 are not part of Pythagorean triple so Sum =1+2 = 3

Euclid's formula is a fundamental formula for generating Pythagorean triples given an arbitrary pair of positive integers m and n with m > n. The formula states that the integers

$a = m^2 - n^2 (1)$

$b = 2mn (2)$

$c = m^2 + n^2 (3)$

form a Pythagorean triple.

From (2) it follows that any even number is valid with the exception of 2 (the smallest is 4, when m=2, n=1) Since very odd integer is the difference of two squares, (1) covers all odd number as valid with the exception of 1. Thus, the solution is 1+2=3.

The two integers are 1 and 2, which sum to 3, the answer. 1 cannot be in a Pythagorean triple, since no two squares sum to 1, nor can 1 be the difference of two squares. Similarly 2 cannot be in a Pythagorean triple, since no two squares sum to 4, nor can 4 be the difference of two squares.

Every other integer is in a Pythagorean triple. To see this, note that, for positive integers a and b with a > b, the integers 2ab, $a^{2}$ - $b^{2}$ , and $a^{2}$ + $b^{2}$ form a Pythagorean triple. I claim that every integer greater than 2 can be obtained from one of these three expressions, for appropriate choice of a and b.

First consider any even integer x greater than 2. Letting a = x/2 and b = 1 in the above expressions, we get a Pythagorean triple containing x.

Now let x be any odd integer greater than 1. I claim we can always find two consecutive squares $a^{2}$ and $b^{2}$ whose difference is x. To see this, compute $i^{2}$ - $(i-1)^{2}$ , which simplifies to 2i-1. This expression generates all odd integers >1 for appropriate choice of i, thus any odd integer greater than 1 can be written as the difference of two squares, and so appears in a Pythagorean triple, using the second of the three above expressions which generate them. This completes the proof.

For any even number the triplet is = 2n , n^2-1 , n^2 +1

For any odd number the triplet is = 2n+1 , 2n(n+1) , 2n(n+1) +1

For 1, 2 we get one of the sides as zero...rest all numbers are possible

Assume that a,b,c are relative prime so a^2 + b^2 = c^2, a^2 = (c-b)(c+b).

If c-b = 1, then a^2 = 2b+1 (a^2 is odd integer).

The smallest possibility for 2b+1 = 3.

In phytagorean, if a^2 + b^2 = c^2, (2a)^2+(2b)^2=(2c)^2.

Conclusion : every number >= 3 can be concluded as phytagorean triplet

From this condition, 1 and 2 are the only remaining numbers, 1+2=3

If n is odd and >1, then n2+(n−12)2=(n+12)2 so n is part of a Pythagorean triplet.

If n=2x, and x is part of a Pythagorean triplet, then so is n=2x since if (p,q,r) is a Pythagorean triplet, then so is (2a, 2b, 2c).

Finally, 4 is part of a Pythagorean triplet since 32+42=52.

Combining the above facts, the only remaining numbers are 1, 2.

The only integers which don't appear are 1 and 2. 3, 4, and 5 appear as part of the smallest primitive Pythagorean triple. From here on, all even numbers are part of a doubled primitive triple, whereas all odds form the shortest legs of new primitive triples (the difference of two consecutive square numbers is always odd).

Thus the answer is 1 + 2, or 3.

For any even number the triplet is = 2n , n^2-1 , n^2 +1 For odd number = 2n+1 , 2n(n+1) , 2n(n+1) +1 For 1,2 we get one of the sides as zero ...rest all numbers are possible Hence sum=1+2=3

Basically, we know that the Pythagorean Triplets of $a^2 + b^2 = c^2$ and that $a^2 = b + c$ , where $a$ is the smaller integer, and $b + 1 = c$ Hence, investigating case-by-case,

Case 1 : $a$ is odd. Thus, $a^2$ is odd, and hence, the value of $b$ and $c$ is always defined. This means that the any odd value of $a$ is always present. Do note that 1 is a special case as to cause the sum of 2 integers to be 1, the only possibility is $0 + 1$ , in which 0 would then not be a positive integer.

Case 2 : $a$ is even. This is actually very unique. Take the form of $3^2 + 4^2 = 5^2$ as an example. To get the even number $6$ , each and every single integer for the base of the squares could be double, and still make logical sense, e.g) $6^2 + 8^2 = 10^2$ This can be proven by that if $a^2 + b^2 = c^2$ is true, then, $(2a)^2 + (2b)^2 = (2c)^2$ , and taking out the $2^2$ , $4a^2 + 4b^2 = 4c^2$ , which simplifies back to the original equation. In addition, recall Case 1, and hence, this means the you can start from an odd number(prime, etc.) and multiply by any integers(most probably an even integer) to obtain the even integer $a$ that you want to prove that it exist. This means that any even integer with product of odd number(s) and the even integer values would hence exist.(Another way to look at it is that it actually means that as long as the prime number 2 is present in the product of the integer's prime factors, this means that it exist.)

So now, what about even integers with prime factors expressed as $2^n$ ? They are also present. Lets reuse Case 1. As 1 cannot be present, the next smallest odd number is 3, and hence, the Pythagorean Triplet $3^2 + 4^2 = 5^2$ . Notice that the integer value 4 is present (2^2), and as this value can be consistently multiplied by 2, this means that for the integers whom prime factors are in the form of $2^n$ , it exists if $n\geq2$ . This also means that 2 is not present in the scenarios given, as a backward derivation from $3^2 + 4^2 = 5^2$ by dividing the triplets by 2 would mean that then, $\frac{3}{2}$ and $\frac{5}{2}$ , in which they are not integers and hence is not applicable.

WIth Case 1 and Case 2 showing that both odd and even numbers with values $>2$ , it would be present in the triplets, hence, the only exceptions are 1 and 2, making the sum of all positive integers which do not appear as one of the entries in a Pythagorean triple, $3$ .

let's consider an integer a a^{2} = c^{2}-b^{2} by Pythagorean Theorem = (c-b)(c+b) for every integer a which is greater than 2, (c-b) can be equal to 1 and (c+b)=a these two equations gives integral values for b and c. All the even integers can appear as multiples of the primitive Pythagorean triples 3-4-5, 5-12-13, 7-24-25, etc. Therefore a is not equal to 1 and 2, 1+2=3

I did it because this result is known to me since I was in VIth standard , but didn't know , Why ?

The submit solution features let me know how it comes Thanks.. brilliant.org , I love you

Every integer n is solution in a Pythagorean triple if $\frac{n^{2}}{2^{k}}$ , where k is any non-negative integer, can be expressed as a sum of two consecutive integers.

We note that the only integer that can't satisfy the above condition, 4, is a part of the triple ( 3, 4, 5), the only ones that are left are 1 and 2.

Therefore the sum should be 3.

We prove that $1$ and $2$ cannot appear in a Pythagorean triple. Neither can be the hypotenuse length because then the leg lengths could not both be positive integers. If they were a leg length, then we would have $1 + b^2 = c^2$ and $4 + b^2=c^2$ respectively. These give $(c-b)(c+b) = 1, 4.$ Because $c-b$ and $c+b$ differ by $2b,$ they have the same parity. Therefore, $c-b = c+b = 1, 2.$ However, this makes $b = 0,$ which is a contradiction. Thus, $1$ and $2$ cannot be in a Pythagorean triple.

Let $a$ be an odd positive integer greater than $1$ . Then $a^2$ is also odd, so $\dfrac{a^2-1}{2}$ and $\dfrac{a^2+1}{2}$ are both positive integers, and furthermore, $\left(a, \dfrac{a^2-1}{2}, \dfrac{a^2+1}{2} \right)$ is a Pythagorean triple. Thus, all odd positive integers except for $1$ appear in a Pythagorean triple.

In addition, multiplying each number in these triples by $2$ produces the numbers that are congruent to $2,$ modulo $4$ (besides $2$ ) as a leg length, so all of those numbers appear in a triple. Finally, multiplying the $3-4-5$ triple by any positive integer produces all of the multiples of $4.$

This shows that all of positive integers besides $1$ and $2$ are in a Pythagorean triple, so the answer is $1+2 = \boxed{3}.$

We know that all odd numbers are the smallest number in some Pythagorean Triple. This is because, ff our odd number is n, the triple will be n, (n^2-1)/2, and (n^2+1)/2. This is because n^2 + ((n^2-1)/2)^2 = n^2 + (n^4-2n^2+1)/4 = 4(n^2)/4 + (n^4-2n^2+1)/4 = (n^4 + 2n^2 + 1)/2 = ((n^2+1)/2)^2.

Remember that 1 does not work here because (1^2-1)/2 = 0, which can not be in a Pythagorean Triple.

Thus, we can rule out all odd numbers over 1.

We also know that every odd number n, where its triple is (n, p, q), has a corresponding triple (2n, 2p, 2q). Each of these numbers can be represented in modular arithmetic as congruent to 2 (mod 4).

Thus, all numbers which are congruent to 2 (mod 4) can be ruled out, starting from 2*3, which is 6. Note that this still leaves 2.

The only case left is the case of a positive integer which is congruent to 0 (mod 4), or a multiple of zero. However, we know that, since (3, 4, 5) is a Pythagorean Triple, and so (3x, 4x, 5x) is a triple for any positive integer value of x. Thus, every multiple of four can be ruled out.

Thus, we have only our two exceptions, 1 and 2, left. 1+2= 3, and 3 is our answer.

Consider the cases of a, where a is the smallest integer in an ordered Pythagorean triple: when a is even and when a is odd. For the even case, keeping in mind that a,b,c are positive integers: a^2+b^2=c^2 --> a^2 = c^2 - b^2 --> a^2 = (c+b)(c-b). If b = c-2, then a^2 = 4(c-1) --> c = ((a^2)/4) + 1. Ordering the triple, a < ((a^2)/4) - 1< ((a^2)/4) + 1. Pyth. Thm. holds true unless a = 2, since b would be 0. Considering a+2 for a, a+2 < ((a+2)^2)/4 -1 < ((a+2)^2)/4 +1. Pyth. Thm. holds true here, too. By induction, a can be all even positive integers except 2. Now the odd case, keeping in mind the aforementioned: a^2 + b^2 = c^2 --> a^2 = (c+b)(c-b). If b=c-1, then a^2=2c-1 --> c = ((a^2)+1)/2. Ordering the triple , a < ((a^2)-1)/2 < ((a^2)+1)/2. Pyth. Thm. is true here unless a =1, since the inequalities would be incorrect. Using induction very similarly to the last case, a can be all odds except 1. Now combining the two cases shows a can be all positive integers except 1 and 2, so the answer is 1+2 = 3

we know that the simplest Pythagorean triplet is 3,4,5. And it's also known to us that we can generate pythagorean triplet by multiplying natural numbers to the mother triplets,Thus we can generate every triplet with any number including.Only 1 and 2 can not be made into included into a triplet...so the those required number is1,2...and the answer is=1+2=3

We know that, all Pythagorean triples are in the form (r^2-s^2)^2 + (2rs)^2 = (r^2+s^2)^2 [Here, we can compare a=(r^2-s^2) , b=2rs , c=(r^2+s^2)], where r and s are positive integers. We know that, the difference between two integer squares is always greater or equal to 3 [when the two integers are 1 and 2, then the difference between them is equal to 3; else, for any other integers, that is greater than 3]. So, (r^2-s^2) \geq 3.When r=s=1, then (2rs)=2; but then (r^2-s^2)=0, which is not possible. So, r \neq s, and if any one among r and s is greater then 1, then (2rs)\geq 3. In this way, we can find that, a and b is always \geq 3. Then c is also \geq 3. And we know that, for every positive integer a or b, we can get a Pythagorean triple. So, there are only two positive integers which are not satisfying the Pythagorean Theorem.Those are 1 and 2, So, the sum is 1+2=3

The Pythagorean theorem can be re-written as $c^2-b^2=a^2$ . By difference of squares we can simplify this to $(c-b)(c+b)=a^2$ . For any odd number $a$ , we can make $b+c=a^2$ and $c-b=1$ to satisfy. Every odd number except 1 can be present in a Pythagorean Triple. Since we have a 3,4,5 triangle, we can multiply 4 by any positive integer. This means that any multiple of 4 can be present in a Pythagorean. To get the multiples of 2 excluding the multiples of 4, we would just use a Pythagorean Triple like 3,4,5 which has an odd number and multiply it by an even number to get all multiples of 2 which are not multiples of 4 except 2. (Since 1 is not present, the argument indicates that 2 is not present either. So, this gives us the answer of $\boxed{3}$

Every integer greater than 2 is part of a primitive or non-primitive Pythagorean triple.

Therefore, the only integers that are not part of a Pythagorean triple are 1 and 2

Sum = 3

General form of pythagorean triplet is 2m,m^2-1,m^2+1.So if we put m=1 then one of the entries would be 0(zero) so no such triangle exists so the answer would be 1 and 2 means 1+2=3

the phythagoras series following this (n, (n^2 -1)/2, (n^2+1)/2 ) in this case just 1 and 2 that we can't use to be n, another integer can.

Pythagorean triples have the form: $m^{2}$ - $n^{2}$ , $2mn$ , $m^{2}$ - $n^{2}$ .

Since $(n+1)^{2}$ - $n^{2}$ = $2n + 1$ , we can assume that every odd positive integer is included as a Pythagorean triple, except 1.

Analyzing the term $2mn$ , we can assume that every even odd positive integer is included as a Pythagorean triple, except 2.

From what is written above, we see that the only positive integers which are not a Pythagorean triple are 1 and 2.

Thus, the sum of all positive integers which do not appear as one of the entries in a Pythagorean triple is 3.

It is well known that Pythagorean triples are generated by the length side of $x^2 - y^2, 2xy, x^2 + y^2$ . Since $(n+1)^2 - n^2 = 2n+1$ , every odd positive integer greater than 1 is included in the entries of Pythagorean triple. By multiplying 2, we also have that every even integer greater than 2 is included. This left 1 and 2 remained. Thus the answer is 1+2=3.

Here is a solution inspired by the parametrization of Pythagorean triples, but it does not strictly rely on it. http://en.wikipedia.org/wiki/Pythagorean triple#Generating a_triple

After a moment of reflection, we see that 1 cannot be part of a Pythagorean triple since the smallest difference of squares of positive integers is $4-1=3$ . So, suppose that $n$ is the $m$ th odd number greater than 1. Then $(m+1)^2 - m^2 = 2m+1 = n$ . Now, notice that $((m+1)^2-m^2)^2 + (2m(m+1))^2 = (m+1)^4 - 2m^2(m+1)^2 + m^4 + 4m^2(m+1)^2 \\= ((m+1)^2 + m^2)^2$ . Thus, $n$ is part of a Pythagorean triple. Finally, suppose that $n$ is an even number greater than 2. Then $n=2m$ for some $m>1$ . We need to have $m-1>0$ for our next computation. Using algebra similar to the above, we see that $(m^2-1)^2 + (2m)^2 = m^4-2m^2+1 + 4m^2 = (m^2 + 1)^2$ . Hence, $n$ is again part of a Pythagorean triple. This leaves only $1+2 = 3$ .

Let a,b,c pythagorean triple , such that : \­( a^2 + b^2 = c^2\­)

So, $b^2 = c^2-a^2 =>$ $b^2 =(c-a)(c+a)$ and if b is a prime number, we have two cases:

$b = c+a$ and $b=c-a$ , but this is impossible because $a \ne 0$ . The second hypothesis is $b^2 = c+a$ and $1 = c-a$ . With this, we have:

$c = \frac{b^2+1}{2}$ and $a = \frac{b^2-1}{2}$ ; and since that $c$ is a integer, the $b$ is odd. Therefore $b$ is an odd and prime number. So, all prime number, except $b=2$ , belongs to pythagorean triple. Since that a compound number is a product of prime numbers, so whole compound number belongs to pythagorean triple too.

Lastly, if $b=1$ , so $c+a = c-a = 1$ (absurd!!).

Therefore, just $1$ and $2$ don't belong a Pythagorean triple. Then, the sum is $1+2 = 3$