Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet? Suppose $a$, $b$ and $c$ are three prime numbers. 
How to prove that $a^2 + b^2 \neq c^2$?
 A: 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 $1^2+1^2=2$.
A: 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 $
A: Hint:

All Pythagorean triples can be written as:
$$  a = k\cdot(m^2 - n^2) ,\ \, b = k\cdot(2mn) ,\ \, c = k\cdot(m^2 + n^2),$$
where $m, n$, and $k$ are positive integers with $m \gt n, m − n$ odd, and with $m$ and $n$ coprime.
A: Hint: If $a,b$ are odd primes, $a^2 + b^2 > 2$ and is even.
Hence, the only possibilities are
$$
2^2 + 2^2 = c^2
\\
2^2 + b^2 = c^2
$$
and they are not possible either because $c - b \ge 2 \implies c^2>2^2 + b^2$.
A: 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. 
A: because one of them must be even number.
A: From $a^2+b^2=c^2$ we get $a^2=c^2-b^2=(c+b)(c-b)$, i.e. a factorization of $a^2$ into two distinct factors $c+b>c-b$. The only such factorizations for the square of a prime is $a^2\cdot 1$, i.e. we conclude $c-b=1$, hence $b=2$, $c=3$. But then $a^2=5$, qea.
