7
$\begingroup$

Let $p$ be a prime so $p\equiv3\pmod4$. If $p| a^2+b^2$, then $p| a,b$

How do I prove this small theorem? I know that it's quite useful. Are there other small theorems like this one? I am mostly searching elementary proofs, so not involving to complicated stuff...

$\endgroup$
  • $\begingroup$ This isn't small. :-) $\endgroup$ – S. Y Sep 16 '16 at 18:51
  • $\begingroup$ Well it isn't recognised as a theorem? Or is it? $\endgroup$ – Taumen Sep 16 '16 at 18:52
  • $\begingroup$ It is a theorem. I will provide a proof later if nobody else does $\endgroup$ – S. Y Sep 16 '16 at 18:55
  • $\begingroup$ Ok Thanks... In french, (I am french ;)) we have something that is called "lemme"... I don't know if that exists in English or something similar, but this one is recognised as a "lemme"... $\endgroup$ – Taumen Sep 16 '16 at 18:57
  • $\begingroup$ Lemme is Lemma in English and German. It comes from the greek word $\lambda\eta\mu\mu\alpha$. $\endgroup$ – Dietrich Burde Sep 16 '16 at 19:01
6
$\begingroup$

I hope I didn't miss something and I think it is pretty elementar:

Using Fermats Little Theorem: $a^p\equiv a\mod{(p)}$ and $b^p\equiv b\mod{(p)}$. Now we get that $a^{p+1}+b^{p+1}\equiv a^2+b^2 \equiv 0 \mod{(p)}$. Because $4\mid p+1$ we can write $p+1=4k$ , for some $k\in\mathbb{N}$. Now we get: $0\equiv a^{4k}+b^{4k}\equiv a^{4k}+(-a^2)^{2k}\equiv a^{4k}+a^{4k}\equiv 2a^{4k} \mod{(p)}$. So now that means $p$ divides $2a^{4k}$, but bcs $p>2$ it cant divide the 2 so it has to divide $a^{4k}$, and if it is a factor of it, it has to be also a factor of $a$, in other words $p\mid a\Rightarrow p\mid b$.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

The ring $ \mathbf Z[i] $ is a principal ideal domain, and any prime that is 3 modulo 4 is inert in this ring. Indeed, writing $ p = (a+bi)(a-bi) = a^2 + b^2 $ and looking at this modulo 4, we find that $ p $ cannot be $ 3 $ modulo $ 4 $. Now, assume that $ p $ divides $ a^2 + b^2 = (a+bi)(a-bi) $, then $ p $ divides one of the factors on the right hand side. Hence, $ p $ divides both $ a $ and $ b $.

Another approach: if we have $ a^2 + b^2 \equiv 0 \pmod{p} $ with $ a, b \neq 0 $, then $ (a/b)^2 \equiv -1 \pmod{p} $, so $ a/b $ has order $ 4 $ in the group $ (\mathbf Z/p \mathbf Z)^{\times} $, which has order $ p - 1 $. This is not divisible by $ 4 $ as $ p \equiv 3 \pmod{4} $, contradicting Lagrange's theorem.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Is there a possibility to prove it without rings and complex numbers? I am searching an elementary proof that is not to complicated... $\endgroup$ – Taumen Sep 16 '16 at 19:00
  • $\begingroup$ But it is a nice proof ;) $\endgroup$ – Taumen Sep 16 '16 at 19:00
  • $\begingroup$ @DanielCortild your theorem means same as if $(a+bi)(a-bi) \cong 0 \mod p$ then $a+bi \cong 0 \mod p$ or $a-bi \cong 0 \mod p$. It's equivalent to Z[i] being a UFD for the special case of primes in Z[i] that are in Z. So there may not be a simpler proof since the algebra of Z[i] really is involved here. $\endgroup$ – djechlin Sep 16 '16 at 19:03
  • $\begingroup$ Ohhh... Ok I will take the time later to sit down and really try to understand it all... But thanks $\endgroup$ – Taumen Sep 16 '16 at 19:05
  • $\begingroup$ I added another, arguably more elementary proof. $\endgroup$ – Ege Erdil Sep 16 '16 at 19:05
1
$\begingroup$

Your assertion can be restated in terms of the quadratic form $q(x,y) = x^2+y^2$ defined over the finite field $\mathbb{F}_p$ of order $p$ (for a prime number $p$): if $p \equiv 3 \pmod{4}$ then for all $(x,y) \in \mathbb{F}^2$, if $q(x,y) = 0$ then $x = y= 0$.

You ask for a generalization, so here is a (useful) one: let $F$ be any field of characteristic different from $2$. For $a,b,c \in F$ consider the binary quadratic form

$q(x,y) = ax^2 + bxy + cy^2$.

We say that $q$ is isotropic if there is $(x,y) \in F^2 \setminus (0,0)$ such that $q(x,y) = 0$ and otherwise anisotropic. And here we go:

(Small but Useful) Theorem: The binary form $q(x,y) = ax^2 + bxy + c y^2$ is isotropic over $F$ if and only if its discriminant $\Delta = b^2-4ac$ is a square in $F$ (meaning $\Delta = d^2$ for some $d \in F$).

Let me sketch the proof: feel free to ask if you want details. Since the characteristic is not $2$, we can diagonalize $q$ just by "completing the square". Moreover, replacing $q$ by $(1/a)*q$ changes the discriminant from $\Delta$ to $\frac{\Delta}{a^2}$ -- so does not affect whether it is a square. So we reduce to the case

$q'(x,y) = x^2 - \frac{\Delta}{4} y^2$, where the result is pretty clear: if $x,y \in F$ are not both $0$ and $q'(x,y) = 0$, then $x \neq 0$ and $y \neq 0$ and $\Delta = (2x/y)^2$. Conversely, if $\Delta = d^2$ then $q'(d/2,1) = 0$.

For the form $q(x,y) = x^2 + y^2$, the discriminant is $-4$, which is a square in $F$ iff $-1$ is a square in $F$. By (very) elementary number theory, when $F = \mathbb{F}_p$ for an odd prime $p$, we have that $-1$ is a square iff $p \equiv 1 \pmod{4}$.

To see why this is useful, now let $a,b,c \in \mathbb{Z}$ and consider the binary quadratic form $q(x,y) = ax^2 + bxy + cy^2$, of discriminant $\Delta$, and suppose that for a prime number $p$ not dividing $\Delta$ we have

$q(x,y) = p$. Then $x$ and $y$ are not both divisible by $p$: if $x = pX$, $y = pY$, then $q(x,y) = p^2 q(X,Y) = p$ is a contradiction. So we find that (the reduction modulo $p$ of) $q(x,y)$ is isotropic over $\mathbb{F}_p$ and thus that $\Delta$ is a square modulo $p$. Using quadratic reciprocity, this translates in every case to congruence conditions on $p$ modulo $\Delta$.

This is really the first step of the arithmetic study of binary quadratic forms over $\mathbb{Z}$. See for instance this lovely book of Cox and these notes based on the book, in particular the first handout. In the latter reference, I call this fact the "fundamental congruence": it appears (in the special case $x^2 + ny^2$) on the very first page of the notes.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.