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This question has already been posted here.

show if $n=4k+3$ is a prime and ${a^2+b^2} \equiv 0 \pmod n$ , then $a \equiv b \equiv 0 \pmod n$

And it also has satisfactory answers however I have not yet studied quadratic residues and though the other answers are well written and I have understood them , my book has given a hint which goes this way :

"If a is not congruent to 0 modulo p then there exists an integer 'c' such that $$ ac\equiv 1\pmod n $$

Any help in approaching this problem using the above hint is appreciated

PS: Please do not mark this as duplicate as I want to understand the usage of this hint in solving the above problem

saisanjeev
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2 Answers2

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You already got one good answer. The following alternative proof is more elementary.

Assume the claim does not hold. Let $n,a,b$ be a counterexample such that $n$ is as small as possible. By the hint, $$(bc)^2+1\equiv(bc)^2+(ac)^2\equiv(a^2+b^2)c^2\equiv 0\pmod{n}. $$ Let $d$ be an even number such that $0<d<n$ and $d\equiv\pm bc\pmod{n}$. Then $d^2+1=mn$ for some positive integer $m<n$. Now $m\equiv 3\pmod{4}$, and therefore $m$ has a prime factor $p$ such that $p\equiv 3\pmod{4}$. Then $d^2+1^2\equiv 0\pmod p$, contradicting the minimality of $n$.

Taneli Huuskonen
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  • At the end of your first step where you have written (bc)^2 + 1 is congruent to 0 mod n can we use the statement that it is solvable only if n is of the form 4k + 1 and conclude the statement? – saisanjeev Feb 16 '18 at 09:54
  • Yes, if you have proved that previously. I included a sketch of the easiest proof I know. – Taneli Huuskonen Feb 16 '18 at 11:09
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Your hint was:

If $a$ is not congruent to $0 \bmod n$ then there exists an integer $c$ such that $ac≡1\bmod n$

-- that is, if $a \not \equiv 0 \bmod n$ then $a$ has a modular inverse.

This would mean that we could take this $c$ and multiply through as follows:

$ a^2+b^2 \equiv 0 \\ 1+(bc)^2\equiv 0 \\ (bc)^2\equiv -1$

meaning that $-1$ would be a quadratic residue$\bmod n$.

We can show this is not the case directly: Since $n$ is prime, there is at least one primitive root such that the order of that number is $\phi(n)=n-1$. And taking one such primitive root $g$, we know that only even powers of $g$ are quadratic residues, but because $-1^2\equiv 1 \bmod n$ we have $g^{(n-1)/2}\equiv -1$ and $(n-1)/2 = 2k+1$ is an odd number, thus $-1$ is not a quadratic residue $\bmod n$.

So $a$ cannot have an inverse and we conclude $a\equiv 0 \bmod p$, with the same for $b$ by symmetry.

Joffan
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