Pythagoras's Prime Triples

We think of parities of $x^2$ , $y^2$ and $z^2$ (if it's odd or even). We claim that there's at least one even term in $x^2+y^2=z^2$ because:

• odd+odd=even

• odd+even=odd

• even+even=even

We know that even*even=even, therefore one term of $(x,y,z)$ is even. The only even prime number is 2. However, we know that there are no Pythagorean triples of positive integers that contain 2. Therefore, our answer is $\boxed{0}$

Using manipulation of the equation by solving for x, we will get

$x^{2} = z^{2} - y^{2} = (z - y)(z + y)$

Which shows that $x^{2}$ do have factors so do $x$ . Another manipulation is by solving for $y$ :

$y^{2} = z^{2} - x^{2} = (z - x)(z + x)$

which also shows that $y^{2}$ do have factors and so do y. Therefore, every Pythagorean triple has at least one prime entry. If x is prime, y must not be prime and vice versa. Therefore the number of Pythagorean triples such that all numbers x, y, and z are prime is 0 .

This is the logic I used, hopefully it's not completely flawed.

I know that the smallest pythagorean triple is 3,4,5. We'll use this later. I also know that an odd times and odd is an odd. Also, I know that an odd plus an odd is an even. Therefore, in order to make a pythagorean triple, the numbers used must consist of either both x and y being odd, and z being even, OR either x or y being even, the other being odd, and z being odd. With all this knowledge, I know that one of the numbers used in a pythagorean triple must be an even number. However, since the only even prime is 2, and we already established that the smallest possible pythagorean triple is 3,4,5, there are no possible pythagorean triples consisting entirely of prime numbers.

for any natural numbers m and n where m and n are relatively prime 2mn , ( $m^2$ - $n^2$ ) and ( $m^2$ + $n^2$ ) will from a Pythhagorean triple. Since the smallest is $3$ , $4$ , $5$ one of the numbers (the 2mn ) will always be even. So there are no such Pythagorean triples where all $3$ are primes. so the ans is $\boxed{0}$

There are no triples that contain a 2. And if all three are odd then we have an odd squared plus an odd squared equals an odd squared which can't be the case because an odd plus an odd is even.

A pythogorian triplet can be expressed in the form of $2m,m^{2}-1,m^{2}+1$ .

The term $2m$ gives us that a triplet can never be prime for any $m$ as $2$ will be one factor in a triplet.

All prime no. ends with 1,3,5(only for 5),7 & 9 (with odd no) except 2 but * 2 is not in any integral triplet of Pythagoras *. (Odd)^{2}=Odd & Odd no +Odd no =even no . But square root of even is always even which can't be prime no. Hence ans is 0.

* If we assume x is odd prime , y is odd prime then we get an even number for z. *

* If we assume x is 2 , y is odd prime then we don't have any possible roots for z. *

* If we assume x is 2 and y is 2 then we don't get possible root for z. *

* Hence there are no such roots. *

Every prime number's square will end in 1 or 9 (Except 2 and 5 which does not form a triple so they are neglected). Possible unit digit for x and y are:

$1+1$ sum ends in 2 which is not possible for a prime's square

$1+9$ sum ends in 0 which is not possible for a prime's square

$9+1$ again not possible for a prime's square

$9+9$ sum ends in 8 not possible for a prime's square

So there are no primes which form pythagorean triples. So the answer is $\boxed0$

We will consider 2 cases.

Case 1: At least one of x, y, z is 2.

Then either $2^{2}+y^{2}=z^{2}$ or $x^{2}+y^{2}=2^{2}.$ Both cases do not have integer solutions.

This is because the first equation implies that $z^{2}-y^{2}=4$ or $(z+y)(z-y)=4.$ Since both $z-y$ and $z+y$ are positive integers, $(z+y, z-y)=(1,4)$ or $(2,2)$ . Both do not give positive integer solutions. The second can be easily verified by trial and error.

Case 2: If all are not 2.

This means all x, y, z are odd. So the squares of x, y and z are also odd. But the sum of an odd number and another odd number cannot be odd, so there are no solutions to this equation.

The answer is 0.

Since 2 is the only even prime number and have no solutions to this problem, every prime number is an odd number. We know that the square of an odd number is also an odd number, so x^2 + y^2 is an even number. So, z^2 is an even number. That's why z is also an even number.

Firstly, 2 cannot make a pythagorean triplet

So if we are considering all x,y,z to be prime ... they have to be all odd numbers

but odd^2 + odd^2 cannot be equal to odd^2 .... hence no solution

We know that all primes except for two are odd. Any odd number squared will result in an odd number. In addition, an odd number plus an odd number is always an even number. Therefore(if x,y,z are primes beside 2) in this equation, the left hand side will be even while the right hand side will be odd resulting in a contradiction. However, we know (from knowing our Pythagorean triplets that no triplets contain 2 in them. Thereby the answer is 0.

Prime numbers are odd with the exception of 2 . Assume, that x,y,z are odd integers. Lets evalute the parity on each side of the expression

$\begin{matrix} odd^2 + odd^2 & = & odd^2 \\ odd + odd & = & odd \phantom{^2} \\ even & \neq & odd \phantom{^2} \end{matrix}$

To end up with an odd parity on the left, one of the integers on the left has to be 2 , which is the only even prime. So our expression becomes:

$2^2 + y^2 = z^2 \quad \Rightarrow \quad 4 + y^2 = z^2 \quad \Rightarrow \quad 4 = z^2 - y^2$

This means the difference between the squares of the two prime number has to be 4. Due to the nature of having to square the integers, y and z have to be relatively small and close in value for the difference to be 4. Also $y \neq z$ since the difference will be 0 .

The only integer solution that works is z = 4 and y = 0 . Neither of which is prime. Furthermore, we can take a look at the difference of the squares for the first couple of primes.

$3^2 - 2^2 = 9 - 4 = 5 \\ 5^2 - 3^2 = 25 - 9 = 16 \\ 7^2 - 5^2 = 49 - 25 = 24 \\ \cdots$

As we can see the larger the integers get, the greater their diffirence. Even our smallest pair of primes, 2 and 3 , does not meet the constraints.

Therefore, the number of Pythogorean triples that meet the requirements is $\fbox{0}$

thinking about parity , as $x,y < z$ ; one of $x , y$ is even (that means only even prime 2). so let y = 2 . $x^{2}=(z+2)(z-2)$ , it leads $z+2 = x^{2}$ and $z-2=1$ or $z=3$ . This all becomes invalid . so there are no $(x,y,z)$ sothat they all could be both pythagorean triple and primes .

Answer is 0. That's cuz, since all have to be prime ( including z ) the sqaures of 2 can't be equal to the square of z

A Pythagorean Triple can never be made up of all odd numbers or two even numbers and one odd number. This is true because:

(i) The square of an odd number is an odd number and the square of an even number is an even number. (ii) The sum of two even numbers is an even number and the sum of an odd number and an even number is in odd number. Therefore, if one of a and b is odd and the other is even, c would have to be odd. Similarly, if both a and b are even, c would be an even number too! and the only even prime is 2...with which we can't make a triple...

$x^2+y^2=z^2$ can be written as $x^2=z^2-y^2=(z+y)(z-y)$ . So, if $x$ is prime, then either $x=z+y=z-y$ or $z-y=1$ . If $x=z+y=z-y$ , then $x=y=z=0$ . So $z-y=1$ . Similarly, $z-x=1$ . This implies that $x=y \implies z^2=2*x^2$ . This implies that z can be only be a prime if $z=2$ . But this would mean that $x=\sqrt{2}$ which is not an integer.

If there is to be a Pythagorean triple of all prime numbers, then all three numbers must be odd, as 2 is not in a pythagorean triple (we can test this easily). So, we have $a^2+b^2=c^2$ , where $a, b,$ and $c$ are all odd numbers. However, since an odd number plus an odd number is even (1 mod 2 + 1 mod 2 = 0 mod 2), we know that this is not possible. Hence the desired answer is $\boxed{0}$ .

At first, we notice 2 can never be part of the askd triple, because 2² is a so small number to complete a squere. Then, we know all others prime numbers are odd. Let's call P=Prime and O=Odd. The question is: P1² + P2² = P3² , and it can be written as: O1² + O2² = O3² . But any odd number squared continue odd, and odd + odd = even Know we have : O + O = O => Even = Odd , and it's abusurd . So, we have 0 triples at all.

If we see in this case all nos. are given prime but while formula for finding out this pythogorean triplets is 2m, m m-1 , m m+1 and we know in this formula there is 2m which means that one no should be composite and not prime , hence answer to this question is 0.

It´s clear that z > 2, since z >= 13, since the smallest possible value of the LHS is 2^2+3^2=13. Therefore, since z is a prime, now established bigger than two, it will be uneven. However, since the z is uneven, the LHS has to have one even number and one uneven number (which summed equal an uneven number), since two unevens = even and two evens=even(even though it´s clear they can´t be both uneven). Therefore, one of the numbers on the LHS must be 2, since it is the only even prime. Therefore, without losing generality, assume y = 2. Therefore, x^2+4=z^2, so (z+x)(z-x)=4. Now, since x,z are both uneven both x+z and x-z will be even. Therefore, the only possible case is z+x=2, z-x=2. Solving this system of equation yields z=2, x=0, which is absurd given the conditions of the problem. Therefore, there are no solutions.