I'm in a pretty simple "CS Math" course for year 1 Comp Sci, and I came across this: Disprove, $x^2 + y^2 = z^2$, such that $x, y, z$ are primes I thought of this as, if n is a prime, then prime factorization of n must be: $n = z*1$ $n^2 = (z*1)(z*1) = (z)(z)$

So I tried to derive a contradiction:

Assume $x, y, z$ are primes such that $x^2 + y^2 = z^2$ Then: $x^2 = z^2 - y^2$

Then: $x^2 = (z-y)(z+y)$ #Contradiction, prime factorization of $x^2$ must be uniquely represented.
But then I noticed that this statement is true if $z-y = 1$ and $z+y = x^2$, So:
Then: $y = -1+z = y = z-1$

Because: $x^2 + y^2 = z^2 $

Then: $(z-1)(z-1) + z+y = z^2$
Then: $z^2 - 2z + 1 + z + y = z^2$
Then: $z^2 - z + y + 1 = z^2$ # subtract $z^2$ from both sides
Then: $-z + y + 1 = 0$
Then: $y = -1 + z$

I remember reducing this to $x + y = z$, which I think is false?

EDIT: Thought of this:
z - y = 1
Then: z = y + 1
Then z must be 3, y must be 2
Because: z is greater than y by 1, since z is a prime and y is a prime, z and y must be 3 and 2 respectively since no other prime number is 1 apart because one of them would be even if greater than 2, and any even number has at least 3 divisors (2, the even number, and 1) thus not a prime.
Then: $x^2 + 2^2 = 3^2$ ,Because $2^2 = 4, 3^2 = 9$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction: x is a prime, but sqrt(5) is not a prime!

Is this a solid evidence proof in that can I say that only prime numbers 1 apart are 2 and 3? I'm pretty sure there is a more simpler and intuitive way to prove this, are there any other ways to prove this via contradiction?

  • $\begingroup$ Your $y=1-z$ should be $y=z-1$, but you square it, so the problem goes away. From $1-z+y=0$ you should get $y=z-1$ so you have nothing new. $\endgroup$ Commented Mar 9, 2015 at 16:54
  • $\begingroup$ Yea, I just noticed that, thanks for pointing it out. How would I prove this then through this? $\endgroup$
    – q.Then
    Commented Mar 9, 2015 at 16:55

3 Answers 3


The easiest way is to note that they cannot all be odd, then show that the even one cannot be $2$.

Added: Let $y=2$. Then $z^2-x^2=(z+x)(z-x)=4$ As $z,x$ are both odd, $z-x \ge 2, z+x\ge 2$, which forces $z+x=2, z-x=2, z=2, x=0$ Contradiction.

  • $\begingroup$ Is there no way of proving this through the method above? $\endgroup$
    – q.Then
    Commented Mar 9, 2015 at 16:59
  • $\begingroup$ @PhilipTsang - within your proof you reach $y=z-1$ - clearly then one of $(y, z)$ is even. $\endgroup$
    – Joffan
    Commented Mar 9, 2015 at 17:04
  • $\begingroup$ Thanks, i actually noticed that as I walking back to my computer. I attached the following evidence above, is that sufficient? $\endgroup$
    – q.Then
    Commented Mar 9, 2015 at 17:14
  • $\begingroup$ Please give enough details to infer how you propose to finish. There are many ways to do this and it is impossible to guess how you propose to continue. $\endgroup$ Commented Mar 9, 2015 at 19:25

If $x,y,z\in\mathbb{N}_{\geq 1}$ and $x^2+y^2=z^2$, then at least one number between $x$ and $y$ is even and at least one number between $x$ and $y$ is a multiple of three, since the only possibilities for a square $\!\!\pmod{3}$ or $\!\!\pmod{4}$ are to be $0$ or $1$. So, in order to prove the statement it is sufficient to check that $2^2+3^2$ is not a square.


I think this proof is sufficient:

If x, y, z are primes, $x^2 + y^2 \ne z^2$
There exists x, y, z that are primes such that $x^2 + y^2 = z^2$, derive contradiction
Then: $x^2 = z^2 - y^2$
Then: $x^2 = (z-y)(z+y)$
Then: $z - y = 1$ and $z + y = x^2$, because by the Fundamental Theorem of Arithmetic, the prime factorization of x which is a prime is:
$x = (x)(1)$
Then: $x^2 = (x)(1)(x)(1) = (x)(x)$
Then: $x^2$ has only three factors, $x^2, 1, x$. Since $x - y \ne x + y$, then by elimination process, $x-y = 1$ and $z + y = x^2$
Then: $x - y = 1$
Then: $x = y + 1$
Then: $x = 3, y = 2$
Because: x and y are primes, x is one greater than y implies that one of x or y must be even as they both cannot be odd (odd number - 1 cannot equal odd). The only even prime number is 2, so by deduction, $y = 2, z = 3$
Also: $z + y = x^2$, as shown above
Then: $x^2 + 2^2 = 3^2$
Then: $x^2 = 9 - 4$
Then: $x = \sqrt5$
Contradiction! Prime Numbers must be natural numbers, $\sqrt5$ is not a whole number or natural number


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