Hi guys, I'm new at number theory. So, Could you help with the following congruence equation.

$\displaystyle $x^2+x+1676 \equiv 0 \mod 2011$$

Regards!

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- Nov 15th 2011, 12:14 PMSoromboMod 2011
Hi guys, I'm new at number theory. So, Could you help with the following congruence equation.

$\displaystyle $x^2+x+1676 \equiv 0 \mod 2011$$

Regards! - Nov 16th 2011, 12:12 PMOpalgRe: Mod 2011
I think that this equation has no solutions. I started by writing it as $\displaystyle x^2 + x - 335\equiv0\pmod{2011}$ (since $\displaystyle 1676\equiv-335\pmod{2011}$). The usual formula for a quadratic equation gives the solutions as $\displaystyle x = \tfrac12(-1\pm\sqrt{1341})$, which you have to write as $\displaystyle x\equiv1006(-1\pm\sqrt{1341})\pmod{2011}$ because 1006 is the inverse of 2 (mod 2011).

So we need to know whether 1341 has a square root (mod 2011). In the technical jargon, is 1341 a quadratic residue (mod 2011)? The easiest way to investigate that is to notice that 1341 is the inverse of 3 (mod 2011). So 1341 will be a quadratic residue if and only if 3 is. By Euler's criterion (which we can apply because 3 and 2011 are both prime), that will be the case if and only if $\displaystyle 3^{1005}\equiv1\pmod{2011}.$ But you can evaluate $\displaystyle 3^{1005}\pmod{2011}$ by writing 1005 = 512+256+128+64+32+8+4+1 and calculating those powers of 3 (mod 2011) by successive squaring and reducing mod 2011. The answer comes out as –1, not 1.

Thus 3 is not a quadratic residue (mod 2011) and therefore that congruence has no solutions.

**Edit**. If you know about quadratic reciprocity, then you can avoid the calculation of $\displaystyle 3^{1005}\pmod{2011}$ and get the result much more painlessly. - Nov 17th 2011, 03:58 PMSoromboRe: Mod 2011
I have a point I'd like to check, how could I prove that x is a quadratic residue modulo p if and only if its inverse is?

- Nov 17th 2011, 11:38 PMOpalgRe: Mod 2011