This is an easy consequence of the fact that if $p \equiv 3$ (mod 4)4) is a rational prime, then $p$ remains prime in the ring of Gaussian integers $\mathbb{Z}[i].$ Then it follows that if $a$ and $b$ are rational integers, the powers of $p$ dividing $a+bi$ and $a-bi$ (in $\mathbb{Z}[i]$ ) are equal. Hence if $n$ is an integer with $a^2 + b^2 = c^2 n$ for integers $a,b$ and $c,$ then the power of $p$ dividing $n$ is even. The fact that $p$ remains prime in $\mathbb{Z}[i]$ was (at least implicitly) known to Fermat. For if $p = (r +si)(t+ui)$ for integers $r,s,t$ and $u,$ then $p^2 = (r^2 +s^2)(t^2 + u^2).$ We can't have $p = r^2 + s^2$ since a sum of two integer squares is congruent to $0,1$ or $2$ (mod 4). Hence we must conclude that one of $r + si$ or $t +ui$ is a unit in $\mathbb{Z}[i]$, so $p$ does remain prime in $\mathbb{Z}[i].$
|
2 | fixed unintended omission | ||
|
|
||||
|
|
1 |
|
||
|
This is an easy consequence of the fact that if $p \equiv 3$ (mod 4), then $p$ remains prime in the ring of Gaussian integers $\mathbb{Z}[i].$ Then it follows that if $a$ and $b$ are rational integers, the powers of $p$ dividing $a+bi$ and $a-bi$ (in $\mathbb{Z}[i]$ ) are equal. Hence if $n$ is an integer with $a^2 + b^2 = c^2 n$ for integers $a,b$ and $c,$ then the power of $p$ dividing $n$ is even. The fact that $p$ remains prime in $\mathbb{Z}[i]$ was (at least implicitly) known to Fermat. For if $p = (r +si)(t+ui)$ for integers $r,s,t$ and $u,$ then $p^2 = (r^2 +s^2)(t^2 + u^2).$ We can't have $p = r^2 + s^2$ since a sum of two integer squares is congruent to $0,1$ or $2$ (mod 4). Hence we must conclude that one of $r + si$ or $t +ui$ is a unit in $\mathbb{Z}[i]$, so $p$ does remain prime in $\mathbb{Z}[i].$ |
||||
