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Consider the numbers $ax^2$, $c-by^2$, where $x,y=0,1,…,p-1$. There are $2p$ such numbers. Any residue class, except for $0$, can have at most two elements of the form $ax^2$ and at most two elements of the form $c-by^2$ from the ones listed above.

I’m trying to understand the logical deductions happening here. Any residue class refers to residue classes of $1$ (mod p), $2$ (mod p),…,$p-1$ (mod p). Then, for residue class $1$ (mod p), there are at most two elements of the form $ax^2$ because both $a(-1)^2$ and $a(1)^2$ are elements of the residue class $1$ (mod p). Is my understanding correct? If my understanding is incorrect, then please tell me what’s happening here. Oh, by the way, $a$ and $b$ are not divisible by $p$.

Chanhyuk Park
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  • $u^2\equiv v^2\pmod p$ if and only if $u\equiv\pm v$. Therefore there are $(p+1)/2$ different residue classes of the form $ax^2$. And also $(p+1)/2$ residue classes of the form $c-by^2$. I guess that is the kind of argument you are striving for. Anyway, this is somewhat of a FAQ. – Jyrki Lahtonen Mar 30 '24 at 19:59
  • See this, and this. – Jyrki Lahtonen Mar 30 '24 at 20:02
  • @JyrkiLahtonen Why are there $(p+1)/2$ different residue classes of the form $ax^2$? – Chanhyuk Park Mar 30 '24 at 20:12
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    @ChanhyukPark From Jyrki's first line it follows that the residues $0^2, 1^2, 2^2, \ldots, [(p-1)/2]^2$ are distinct mod $p$. How many different values is that? – Erick Wong Mar 30 '24 at 20:18

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