75 of 100: Odd Triples

Proof without words:

$\big({\color{#D61F06}{2m + 1}}\big)^2 + \left(2m^2 + 2m\right)^2 = \left(2m^2 + 2m + 1\right)^2$

We notice the following things: $3^2=9=(5+4)(5-4)=5^2-4^2$ $5^2=25=(13+12)(13-12)=13^2-12^2$ $7^2=49=(25+24)(25-24)=25^2-24^2$

Now, we try to generalize this. $(2n+1)^2=4n^2+4n+1=(\frac{4n^2+4n}{2}+1+\frac{4n^2+4n}{2})(\frac{4n^2+4n}{2}+1-\frac{4n^2+4n}{2})$ $=(\frac{4n^2+4n}{2}+1)^2-(\frac{4n^2+4n}{2})^2$ $=(2n^2+2n+1)^2-(2n^2+2n)^2$

And, two consecutive integers are coprime. So, this triplet is primitive.

P.S. Maybe I need to clarify my generalization process a bit more. Take $49$ . We'll write it as $(\frac{48}{2}+1)+\frac{48}{2}$ . That's what I did with $4n^2+4n+1$ . I wrote it as $(\frac{4n^2+4n}{2}+1)+\frac{4n^2+4n}{2}$ .

We know that the difference between 2 consecutive square quantities is odd number.
So,. (n+1)²—n²=(2k+1).
Now, the square of an odd number is a odd number.i.e. in the form of (2k+1).
It is obvious that, g.c.d. of (n , n+1)=1.
So, the numbers are primitive.
Hence, the statement is true.

All primitive Pythagorean triplets can be generated by the formula $a = p^2 - q^2,\ \ \ b = 2pq,\ \ \ , c = p^2 + q^2,$ with $p, q$ coprime and not both odd.

In order to make $a = 2n+1$ an arbitrary odd number, we choose $p = n$ and $q = n+1$ . These are obviously coprime and not both odd. Then $a = p^2 - q^2 = (p+q)(p-q) = (2n+1)\cdot 1 = 2n+1$ as desired.

Suppose a is an odd positive integer.

Then to be the smallest member of a Pythagorean triple, we need b and c such that $a^2 = c^2 - b^2 = (c-b)(c+b)$ .

Take $c = b+1$ , then $\dfrac{a^2 -1}{2} = b$ . Because a is odd, $b \in \mathbb{N}$ .

So a,b,c form a Pythagorean triple and a is the smallest member. Since $c= b+1$ , is it also primitive.

We can test this with as many odd numbers as we like. For example, 11^2 = 121, and the two consecutive numbers that add up to 121 are 60 and 61, and 11^2+60^2 = 61^2. 13^2 = 84+85, and 13^2+84^2 = 85^2

Answered in two words: Of course!

Let the smallest side be 2n + 1 (n > 0) and the other sides be y and z, with

the largest side being z, then

z^2 = y^2 + (2n + 1)^2, (1)

Then obviously y and z have different parity. Then

(z – y)(z + y) = (2n + 1)^2

Since the gcd(z – y, z + y) | 2 gcd(z, y) = 2 and z – y and z + y are odd,

so gcd(z – y, z + y) = 1. So, (1) shows that z – y = (2m + 1)^2, z + y = (2k + 1)*2.

z = 2k(k + 1) + 2m(m + 1) + 1, y = 2k(k + 1) – 2m(m + 1) and

(2m + 1)(2k + 1) = (2n + 1). If 2n + 1 is any odd number, we can set, 2k + 1 = 2n + 1,

2m + 1 = 1,So, z = 2n(n + 1) + 1 and y = 2n(n + 1). So we have:

[2n(n + 1) + 1]^2 = [2n(n + 1)]^2 + (2n + 1)^2

This is actually what we are looking for.

Let d = gcd (z, 2n + 1), obviously d | (z – 2n – 1) = 2n^2. Since (2n + 1, 2n^2) = 1,

then d = 1. So obviously 2n + 1, y, and z are a primitive triple.

In general case 2n + 1, factor it as (2k + 1)(2m + 1), such that k > m, and

(2k + 1, 2m + 1) = 1, then z + y = (2k + 1)^2, z - y = (2m + 1)^2.

In this case, z = 2k(k + 1) + 2m(m + 1) + 1, and y = 2k(k + 1) - 2m(m + 1)

[2k(k + 1) + 2m(m + 1) +1]^2 = [2k(k + 1) - 2m(m + 1))^2 + (2n + 1)^2

Since k < n, then y < 2n(n + 1), which shows y < 2n + 1, so in this case 2n + 1 is not the smallest

leg, so we have to choose the first approach as the third example below shows.

1) Let 2n + 1 = 13, then k = n = 6, m = 0, z = 85, y = 84

85^2 = 84^2 + 132, 7225 = 7056 + 169

2) Let 2n + 1 = 81, then k = 40, m = 0, z = 3281, y = 3280

3281^2 = 3280^2 + 81^2, 10764961 = 10758400 + 6561

3) Let 2n + 1 = 63, then k = 4, m = 3, z = 65, y = 16

65^2 = 16^2 + 63^2, 4225 = 256 + 3969

Although this is primitive triplet, it is not what we are looking for.

So, we should let k = 31, m = 0, then z = 1985, y = 1984,

1985^2 = 1984^2 + 632 3940225 = 3936256 + 3969.

This is what we are looking for.

Every Pythagorean triplet is of the form

$2kmn, k(m^{2}-n^{2})$ and $k(m^{2}+n^{2})$

Allow $k$ to be 1 and then the second member is

$(m+n)(m-n)=O$ for any odd integer $O$ .

One way can be

$m+n=O$ and $m-n=1$ . There always exists such integers $m$ and $n$ , right?

[We can also show that they are primitive.]

Notice that the large sides are always b and b+1. Solve:

(2a+1)^2 + b^2 = (b+1)^2

Reduces to 2a^2 + 2a = b

So for each value of a there is an odd short side (2a +1) and two long sides length 2a^2 + 2a and 2a^2 + 2a +1.

Using Euler's formula for constructing Pythagorean triples, instead of using two distinct values $m$ and $n$ , if we instead use $m$ and $m + 1$ for all the natural numbers, we will see that the first term $n^2 - m^2$ will simplify to $2m + 1$ which is the general term for all the odd natural numbers.

Furthermore, subbing in $m$ and $m + 1$ into the other two values of $2mn$ and $n^2 + m^2$ , we get two terms with a value of $2m^2 + 2m$ and $2m^2 + 2m +1$ , which are two consecutive natural numbers whose sum is equivalent to $(2m +1)^2$ , therefore always making the value of the other two sides of the primitive triple add up to the square of that shortest side.

Another way of looking at this is by choosing any natural number $m$ and then making substitutions of $a = 2m+1$ and $b=2m^2 + 2m$ , we create Pythagorean triples of the form $a, b, b+1$ , where $a^2 = 2b + 1$ .

Here is an interesting way to find the larger two members of a Pythagorean triple of any odd number a, a >1 in terms of the sums of odd numbers. (Still working on the primitive part for this one).

Here is a complete proof. Given any odd number greater than 1 a primitive Pythagorean triple can be derived directly as follows:

Take the nth odd number $A = 2n + 1$ n = 1,2,3,...

$A^2 = (2n + 1)^2 = 4n^2 + 4n + 1$

Set $B = n A + n = n (2n + 1) + n = 2n^2 + 2n = 2n(n+1)$

Set C = B + 1

$C^2 = (B + 1)^2 = B^2 + 2B + 1$

$C^2 - B^2 = 2B + 1 = 2(2n(n+1)) = 4n(n+1) + 1 = 4n^2 + 4n + 1 = A^2$

Therefore for any odd number, A, there are values B and C such that $C^2 = A^2 + B^2$ , forming a Pythagorean triple.

A and B have no common factors therefore it is also a primitive Pythagorean triple.

Does not really draw much blood. The consecutive squares in the given examples provide an excellent clue that suggests the proof above.

For any odd number n, you can calculate n^2, then find a such that 2a+1=n^2. Once you have this, you can write the expression n^2+a^2, and we said that n^2=2a+1, so the expression comes out to be a^2+2a+1, which factors to (a+1)(a+1), or just (a+1)^2. Thus, we have found a Pythagorean triple.

Let a = 2m + 1 be an arbitrary odd positive integer. Let b = 2m(m + 1), and c = m^2 + (m + 1)^2. Then a^2 + b^2 = c^2 is clearly a primitive Pythagorean Triplet. Ed Gray

Let $1<a<b<c$ , and let $a$ be odd as per the statement. The Pythagorean statement can be stated as $a^2=c^2-b^2$ . Since $a$ is odd, $a^2>1$ is guaranteed to be odd as well.

Notice the pattern between pairs of consecutive squares: $(n+1)^2-n^2=2n+1$ , which is also always odd. Obviously, $a^2=2n+1$ for some positive $n$ . This means that we can say $a^2=(\frac{a^2-1}{2}+1)^2-(\frac{a^2-1}{2})^2$ , forming the Pythagorean triple $a^2,\frac{a^2-1}{2},\frac{a^2-1}{2}+1$ where $a^2$ is both odd and the smallest member of such triple.

However, we have not yet shown that the Pythagorean triple is primitive, not sharing a factor. If just two members are relatively prime, the entire triple must also be and thus primitive. Let's examine $\frac{a^2-1}{2}$ and $\frac{a^2-1}{2}+1$ . Adjacent numbers are always relatively prime (try to think of a number that divides both). We are now guaranteed the Pythagorean triple is primitive. Q.E.D.

Lemma: every odd number greater than 1 can be represented by (x^2 - (x-1)^2.)

Proof: $x^2 - (x-1)^2 = x^2 - (x^2-2x+1) = 2x - 1$

Using the equation $a^2+b^2=c^2$ , let b equal c-1 and subtracting it to get $a^2 = c^2 - (c-1)^2.$ Notice the RHS can be any odd number by the Lemma above. Since a^2 is odd when a is odd, we can represent it with sides c and b. The triple must be primitive because gcd(a,c-1,c)>=gcd(c,c-1)=1.

I remember there is a formula for Pythagorean triple

x=2*k+1

y=2 k (k+1)

z=(k+1)^2 +k^2

Where k is some natural number and x^2+y^2=z^2

So from those formula is clear than the smallest member is x, which is an odd number

think about it people ( its 3)

n^2 + x = (n+1)^2 => x = 2n +1 so yeah not only "Every odd number greater than 1 is the smallest member of a primitive Pythagorean triple.", but the other 2 numbers are consecutive one's.

We can represent all odd numbers greater than 1 as $1+2x \forall x \in \mathbb{N}^+$ So, using the pythagorean theorem, we get $(1+2x)^2+a^2=b^2 \\ (1+2x)^2=b^2-a^2 \\ (1+2x)^2=(b-a)(b+a)$ Because $1+2x$ is odd, either $a$ or $b$ is odd and the other is even. Let's assume $a$ is even, so we can represent $a$ as $2y$ , so we get $(1+2x)^2=(b+2y)(b-2y)$ . Based on the calculated pythagorean triples given to us, let's assume that $b=a+1=2y+1$ . Now, plugging this into our equation, we get $(1+2x)^2=(2y+2y+1)(2y+1-2y) \\ (1+2x)^2=(4y+1)(1)=4y+1$ Calculating this out, we get $4x^2+4x+1=4y+1 \\ 4(x^2+x)+1=4y+1 \implies y=x^2+x$ Because $x$ is a positive natural number and $2y=2x^2+2x>x$ , we now know that for all odd numbers greater than 1, it is the smallest member of a pythagorean triple.

...for every odd integer j, there exist infinitely many primitive Pythagorean triples in which the hypotenuse and the even leg differ by j^2. https://en.wikipedia.org/wiki/Pythagorean_triple

For any odd a, b = (a^2-1)/2 and c = b+1 form a Pythagorean triple, and as long as a>2 (which is true for all odd numbers >1), a<b<c. b and c are obviously relatively prime, and it is possible to show that a is relatively prime with both b and c.

x^2 + y^2 = z^2. Now write that differently, so x^2 = z^2 - y^2. Now we'll think only of a Pythagorean triple with the largest number only 1 greater than the 2nd number. So z = y + 1. So x^2 = (y + 1)^2 - y^2. Hence x^2 = (y^2 + 2y + 1) - y^2. Hence x^2 = 2y + 1 - which is of course odd. Hence 2y = x^2 - 1. Hence y = 1 / 2 x (x^2 - 1), And then z is one more than that. So z = 1 / 2 x (x^2 + 1). So for any value of x>1 (because when x=1, y=0 and then z=1 so this is not a triangle) y and z can be found. So for example when x=19 y = 1 / 2 (19^2-1) which is 180 and then z is 1 more than it, so z = 181.