Find all positive integers $x$ such that $13 \mid (x^2 + 1)$

I was able to solve this by hand to get $x = 5$ and $x =8$. I didn't know if there were more solutions, so I just verified it by WolframAlpha. I set up the congruence relation $x^2 \equiv -1 \mod13$ and just literally just multiplied out. This lead me to two questions:

1. But I was wondering how would I do this if the $x$'s were really large? It doesn't seem like multiplying out by hand could be the only possible method.

2. Further, what if there were 15 or 100 of these $x$'s? How do I know when to stop?

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If $p$ is an odd prime, and $a$ is not divisible by $p$, then the congruence $x^2\equiv a\pmod{p}$ has $0$ or $2$ solutions modulo $p$. You have found two incongruent solutions. So you have all of them: all solutions are of the form $x=5+13k$ or $x=8+13k$, where $k$ ranges over the integers.

Actually, finding one solution would be enough, for if $x$ is a solution, automatically so is $-x$.

For prime $p$, there are good algorithms for computing solutions of $x^2\equiv a \pmod{p}$, that are feasible even for enormous $p$.

If the modulus is not prime, things get more complicated. Suppose that $m$ is an odd number $\gt 1$. Let the number of distinct prime divisors of $m$ be $e$. Then the congruence $x^2\equiv a\pmod{m}$, where $a$ and $m$ are relatively prime, either has $0$ solutions or $2^e$ solutions.

Finding the solutions can be computationally difficult. If $m$ is the product of two distinct primes, then finding the solutions is essentially equivalent to factoring $m$. This is believed to be in general computionally very difficult for enormous $m$.

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Is there a general form of the first sentence you stated? – AlanH Jun 2 '13 at 4:53
One can get generalization to $x^k\equiv a\pmod{p}$, criterion for existence of solution, a count on the number of solutions. Not as well developed as the theory of quadratic residues, but pretty good. Might google power residues, indices. – André Nicolas Jun 2 '13 at 5:10

Starting with $2,$ the minimum natural number $>1$ co-prime with $13,$

$2^1=2,2^2=4,2^3=8,2^4=16\equiv3,2^5=32\equiv6,2^6=64\equiv-1\pmod{13}$

As $2^6=(2^3)^2,$ so $2^3=8$ is a solution of $x^2\equiv-1\pmod{13}$

Now, observe that $x^2\equiv a\pmod m\iff (-x)^2\equiv a$

So, $8^2\equiv-1\pmod {13}\iff(-8)^2\equiv-1$

Now, $-8\equiv5\pmod{13}$

If we need $x^2\equiv-1\pmod m$ where integer $m=\prod p_i^{r_i}$ where $p_i$s are distinct primes and $p_i\equiv1\pmod 4$ for each $i$ (Proof)

$\implies x^2\equiv-1\pmod {p_i^{r_i}}$

Applying Discrete logarithm with respect to any primitive root $g\pmod {p_i^{r_i}},$

$2ind_gx\equiv \frac{\phi(p_i^{r_i})}2 \pmod {\phi(p_i^{r_i})}$

as if $y\equiv-1\pmod {p_i^{r_i}}\implies y^2\equiv1$ $\implies 2ind_gy\equiv0 \pmod {\phi(p_i^{r_i})}\implies ind_gy\equiv \frac{\phi(p_i^{r_i})}2 \pmod {\phi(p_i^{r_i})}$ as $y\not\equiv0\pmod { {\phi(p_i^{r_i})}}$

Now apply CRT, for relatively prime moduli $p_i^{r_i}$

For example, if $m=13, \phi(13)=12$ and $2$ is a primitive root of $13$

So, $2ind_2x\equiv 6\pmod {12}\implies ind_2x=3\pmod 6$

$\implies x=2^3\equiv8\pmod{13}$ and $x=2^9=2^6\cdot2^3\equiv(-1)8\equiv-8\equiv5\pmod{13}$

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We can do this avoiding the congruence machinery,although it is essentially the same idea. Let x = 13a + b, where 0<=b<=12. Then x^2 +1 = 169a^2 + 26ab + b^2 +1.You only need to find values of b such that b^2 + 1 is divisible by 13.This is a very small sample,as you found that b = 5 or b =8.Then x =13a +5 or x = 13a +8,and you have infinitely many answers. Edwin Gray

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