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Find all solution of $x^2 + 1 \equiv 0$ (mod 103459).

103459 is the product of two primes 307 and 337.

What I do is to rewrite the equation as:

$x^2 + 1 \equiv 0$ (mod 307)

$x^2 + 1 \equiv 0$ (mod 337)

Then we have:

$x^2 + 1 = 307m$

$x^2 + 1 = 337n$

and hence $307m = 337n$ $=>$ $m=337, n=307$

$x^2 = 307m - 1 = 103459 - 1 = 103458$

I stop here because it I feel it's not the way to do it.

Is there any better way to solve this question?

yashirq
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  • See http://math.stackexchange.com/questions/122048/1-is-a-quadratic-residue-modulo-p-if-and-only-if-p-equiv-1-pmod4 – lab bhattacharjee Nov 03 '16 at 04:52
  • For the solution you tried: when you look for solutions of $x^2+1\equiv0\pmod{307}$ and $x^2+1\equiv0\pmod{337}$, you called both hypothetical solutions $x$; that's a mistake, as they will probably be different numbers. If you gave them different names as appropriate, you would avoid the trap of equating the two expressions. – Greg Martin Nov 03 '16 at 05:38

1 Answers1

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It turns out there are no solutions. $307 \equiv 3 \pmod{4}$ and primes of the shape $4k+3$ don't have $-1$ as a quadratic residue. So there's no solution to $x^2\equiv -1 \pmod{307}$. The other congruence has a solution (two solutions, really) but it's tedious to find them, and, at this point, not useful.

B. Goddard
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