3
$\begingroup$

It is easy to find dozens of proofs (some of which are elementary or easy to follow) that every prime of the form $4m+1$ can be expressed as the sum of two squares.

How can one prove that every prime of the form $4m+1$ can be expressed as the sum of two squares $p = a^2 + b^2 | a,b \in \Bbb Z^+, a>b$ in only one way?

This is easy to verify for primes up to quite a large value but I can't find a proof.

$\endgroup$
  • $\begingroup$ Most elementary texts on number theory prove this. $\endgroup$ – Lord Shark the Unknown Nov 20 '18 at 20:28
  • $\begingroup$ @mathnoob $3$ is not of the from $4m+1$ $\endgroup$ – saulspatz Nov 20 '18 at 20:38
  • 3
    $\begingroup$ very nice discussion of this, also for $mx^2 + ny^2$ with $m,n$ positive and coprime, by Brillhart zakuski.utsa.edu/~jagy/Brillhart_Euler_factoring_2009.pdf $\endgroup$ – Will Jagy Nov 20 '18 at 20:43
  • $\begingroup$ @Will Jagy - Thanks for that reference - very enlightening. $\endgroup$ – marty cohen Nov 20 '18 at 20:52
  • $\begingroup$ @will Jagy yes that answers the question. I have +1 the comment but if you put the proof as an answer it will bury this question; otherwise, perhaps I will do so maybe tomorrow. $\endgroup$ – Mark Fischler Nov 20 '18 at 20:57
2
$\begingroup$

Denote by $p$ a prime number of the form $4m+1$. Thus $$p\equiv 1\pmod 4$$ Assume that $p={a^2}+{b^2}={c^2}+{d^2}$ for some $a,b,c,d\in\mathbb {N}$ such that, w.l.o.g. $a>b$ and $c>d$. Our aim is now to show by contradiction, that this follows to $a=c$ and $b=d$:

We now have $$(ad-bc)(ad+bc)={a^2}{d^2}-{b^2}{c^2}=\bigl(p-{b^2\bigr)}{d^2}-{b^2}\bigl(p-{d^2\bigr)}=p\bigl({d^2}-{b^2}\bigr)\equiv 0\pmod p$$ This implies (see Euclid's lemma) that either $$p|(ad-bc)$$ or $$p|(ad+bc)$$

Let's suppose $p\mid(ad+bc)$. Hence ${a^2},{b^2},{c^2},{d^2}<p$. Thus $a,b,c,d<\sqrt{p}$, which implies that $$0<ad+bc<2p$$ That leads to $ad+bc=p$. However $${p^2}=\Bigl({a^2}+{b^2}\Bigr)\Bigl({c^2}+{d^2}\Bigr)={(ad+bc)^2}+{(ac-bd)^2}={p^2}+{(ac-bd)^2}$$ $$\Rightarrow ac-bd=0$$ which, nevertheless, contradicts the assumption that $a>b$ and $c>d$. This shows, that $p\nmid ({ad+bc})$.

This implies $p\mid ({ad-bc})$. Analugously $$a,b,c,d<\sqrt{p}$$ which implies that $$-p<ad-bc<p \Rightarrow ad=bc$$ Hence $a\mid {bc}$. Note however, that since $p={a^2}+{b^2}$, $a$ and $b$ have to be coprime, so $$a\mid{c}$$ Let $c=xa$. Since $ad=bc$, it follows that $d=xb$

We finally have $$p={a^2}+{b^2}={c^2}+{d^2}={x^2}\bigl({a^2}+{b^2}\bigr)\Rightarrow x=1$$ $$\Rightarrow c=a\quad d=b$$

$\endgroup$

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.