Suppose $a$, $b$ and $c$ are three prime numbers.
How to prove that $a^2 + b^2 \neq c^2$?
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The sum of two odd numbers are even, so one of the numbers must be $2$.
If $a$ or $b$ are $2$ we have $a^2+4=c^2$ or $4=(c+a)(c-a)$ Since $c-a$ and $c+a$ have the same parity, this is impossible.
If $c=2$ we have $a^2+b^2=4$ but since $a$ and $b$ are positive, both must be $1$, but $1^2+1^2=2$.
Primes are all odd expect $2$, so if $a, b, c$ don't contain $2$, $a^2 + b^2$ is even but $c^2$ is odd, then $a^2 + b^2 = c^2$ can't be true.
Of course if $c=2$, then $a^2 + b^2 = c^2$ can't be true.
If $a = 2$, then $c>b$ then $c^2 - b^2 \geq (b+2)^2 - b^2 = 4b + 4 > a^2 $
On cannot have $a^2+b^2=c^2$ if $a,b,c$ are all odd. Since $2$ is the only even prime number, one of $a,b,c$ would have to be $2$. But since $2$ is the smallest prime number clearly $c$ cannot be$~2$; furthermore for any $n\geq2$ one has the inequalities $n^2<n^2+4<(n+1)^2$ showing that $n^2+2^2$ is not a square, so $a$ or $b$ cannot be $2$ either.