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So I had to find the condition (for $a$, $b$, $p$; $p$ is prime) when $$ (condition) \iff (a^2+b^2 \equiv 0\mod p \Rightarrow a^2+b^2 \equiv 0 \mod p^2)$$ is true.

When $a \equiv 0\pmod p$ and $b \equiv 0\pmod p$ it is always true, but there are non-trivial cases when $a \neq 0\pmod p$ and $b \neq 0\pmod p$ and the implication is correct (ex. $a = 3, b = 4, p = 5$)

I tried to express $a^2+b^2 \equiv 0\pmod p$ as: $$(q_1*p + r_1)^2 + (q_2*p + r_2)^2 \equiv 0 \pmod p$$ where $r_1$ and $r_2$ in $[-\frac{p-1}{2}; \frac{p-1}{2}]$ to get: $$p^2(q_1^2+q_2^2) + 2p(q_1*r_1 + q_2*r_2) + (r_1^2+r_2^2) \equiv 0 \pmod p$$ The only thing I realised is that $r_1^2 \equiv -r_2^2 \pmod p$ and I don't know what to do next.

Also, it may be useful: for every $p= 4k+3$ this implication works because for these primes a and b can only be congurent to $0$ mod $p$ and every $p=4k+1$ it is true that: $a^2+b^2 = p$ has a unique soultion with $a\le b$ but I have not managed to do sth with it yet(

Bill Dubuque
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Artkol
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  • Not following. $3^2+4^2=25$, for example, so there are non-trivial solutions $\pmod {p^2}$ for some $p$. – lulu Apr 27 '24 at 12:14
  • Hensel's lemma lets you take a solution mod $p$ and lift to mod $p^2$ to satisfy it, but it won't always be the case. In particular, suppose $a^2+b^2=0\mod p$ then you can look at $(a+cp)^2+(b+dp)^2=0 \mod p^2$ and you end up with $\frac{a^2+b^2}{p}+2ac+2bd=0 \mod p$ which lets you solve for $c$ in terms of $d$ which has $p$ choices. So "on average" this is only true $\frac{1}{p}$ of the time. – Merosity Apr 27 '24 at 12:15
  • @lulu that's the thing, there are non-trivial solutions and I don't know whether it is possible to set some common condition for all non-trivial cases. All I know is that non-trivial solutions appear only for p=4k+1 – Artkol Apr 27 '24 at 12:24
  • Oh sorry, I made a mistake when I posted, yeah, there are non-trivial cases, ty! – Artkol Apr 27 '24 at 12:31
  • What do you mean by the condition? You could just put the same statement on both sides of the equivalence and be done. But surely that's not what you're after...? – Servaes Apr 27 '24 at 17:24
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    @Servaes The condition should be some convenient criteria for a,b and p that it could be possible to quickly determine whether the implication is correct. E.x. if $p=4k+3$ or $p=4k+1$ and $a^2≡−1 \mod p$ ⟺ $a^2+b^2$ ≡ $0 \mod p$ ⇒ $a^2+b^2$ ≡ $0\mod p^2$ (this is just illustration) – Artkol Apr 27 '24 at 19:07

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