Prime numbers qns

Question

Show that for any prime number $p$, $q$, $r$, one has $p^2+q^2$ does not equal to $r^2$.

I have no idea how to start and prove it. One stumbling part is that we cannot deduce with certainty that r must be odd. I tried using the $(p+q)^2$ identity but cannot proceed further.

Answer 1

Look at the equation $\pmod 4$. $p,q$ cannot be both odd since $2$ is not a quadratic residue $\pmod 4$. So say $p$ is even so $p = 2$. Then look at the equation $\pmod 3$. $q^2$ is either $1$ or $0$ $\pmod 3$ but it cannot be $1$ so it must be $0 \pmod 3$. Thus $q = 3$ and $2+3 = 7$ is not a square.

Answer 2

The sum of squares of two odd primes cannot be a square.

Indeed an odd prime is congruent to $\pm 1\mod 4$, so that the sum of their squares is congruent to $2$, which is not a square modulo $4$.

Thus if $p^2+q^2=r^2$, one of $p,q$, say $p$, must be $2$, clearly not both, so we can write $r^2=4+q^2$. However, modulo $8$, all odd squares are congruent to $1$, and $4+q^2\equiv 5\mod 8$, which is not an odd square modulo $8$.

Answer 3

$p^2 = r^2 - q^2 = (r-q)(r+q)$, and $r+q > r - q$. Therefore we must have $r+q = p^2$ and $r - q = 1$. But the only primes that differ by $1$ are $2$ and $3$, and $3+2 = 5$ is not a square.