Find all solutions of $x^2 \equiv 9 \pmod{85}$

Question

I am asked to solve this problem, and I know how to solve congruences of degree $2$ modulo a prime $p$, but note that $85=5\cdot 17$ is a product of two primes.

On the fly I managed to rewrite the expression as $(x-3)(x+3)\equiv 0 \pmod{85}$, and then since $(5,17)=1$, $5\cdot 17 | (x-3)(x+3)$ implies $5|(x-3)(x+3)$ and $17|(x-3)(x+3)$.

Thus I tried expanding to two simultaneous congruences where the moduli are prime:

$$x^2 \equiv 9 \equiv 4 \pmod{5}$$ $$x^2 \equiv 9 \pmod{17}$$

which I now know how to solve, for instance, the first congruence can be done by inspection as $x\equiv 2,3$ are the only such solutions (there are only four to check). I further conclude that $x=2+5k,3+5k'$ for $k,k' \in \mathbb{Z}$, and plug these values into the second equation:

$$(2+5k)^2 \equiv 4+10k+10k+25k^2 \equiv 4+20k+25k^2 \equiv 4+3k+8k^2 \equiv 9 \pmod{17}$$ $$(3+5k')^2 \equiv 9+15k'+15k'+25k'^2 \equiv 9+30k'+25k'^2 \equiv 9+13k'+8k'^2 \equiv 9 \pmod{17}$$

thus $8k^2+3k-5\equiv 0 \pmod{17}$ and $8k'^2+13k' \equiv 0 \pmod{17}$ where $d=b^2-4ac = (3)^2-4(8)(-5)=13^2$ and $d'=b^2-4ac = 13^2-4(8)(0)=13^2$ thus from my book it states that both are solvable and with solutions:

$$k = \frac{-3\pm 13}{16}$$ $$k'=\frac{-13 \pm 13}{16}$$

where $\frac{1}{16}=16$ is the inverse of $16$ modulo $17$ thus we get:

$k\equiv (-3+13)\cdot 16 \equiv 10\cdot 16 \equiv 160 \equiv 7 \pmod{17}$

$k \equiv (-3-13)\cdot 16 \equiv -16\cdot 16 \equiv -(16)^2 \equiv -1 \equiv 16 \pmod{17}$

$k'\equiv (-13+13)\cdot 16 \equiv 0\cdot 16 \equiv 0 \pmod{17}$

$k' \equiv (-13-13)\cdot 16 \equiv -26\cdot 16 \equiv -8 \equiv 9 \pmod{17}$

and finally $2+5(7+17\ell) =37 +85\ell, \:\:2+5(16+17\ell) =82 +85\ell,\:\:3+5(0+17\ell) =3 +85\ell,\:\:3+5(9+17\ell) =48 +85\ell$ for $\ell \in\mathbb{Z}$.

$x \equiv 3,37,48,82\pmod{85}$. I ran a computer check and it gave me the same four numbers so I guess it's correct.

Is my approach correct?

Maybe I was lucky with this method for this particular problem, but I'm wondering when would my approach break down?

Because I've never seen congruences of degree $2$ where the modulus is not prime, let alone powers or multiples of primes.

Answer 1

Your method is absolutely correct, and it works generically under the name of "Chinese Remainder Theorem". Essentially, the result says that if $m_1,\ldots,m_t$ are pairwise coprime then a system of equations of the form

$$\begin{cases} x= a_1 \mod m_1 \\ \hphantom{W}\vdots \\ x= a_t \mod m_t \end{cases}$$

has a unique solution modulo $m = m_1\cdots m_t$. The algorithm you followed to find solutions is indeed correct, and is one way of approaching your problem.

Another quicker way goes as follows:

1. from $x^2=9\mod 17$ you get that $x=3$ or that $x=-3=14 \mod 17$, since $17$ is prime.
2. from $x^2=4\mod 5$ you get that $x=2$ or that $x=-2=3\mod 5$, since again $5$ is prime.

Now you need to find $x$ such that $(x\mod 5,x\mod 17) = (a,b)$ where $(a,b)$ runs through the four possible pairs you have obtained. To do this, you can note that the inverse of $5$ modulo $17$ is $7$, and hence all answers are obtained by computing

$$a+5\cdot 7\cdot(b-a) \mod 85.$$

This indeed what you obtained.