# Find all solutions for $7x^2 \equiv 3 \mod5$, if any.

First of all in $\mod5$, the congruence could be reduced to $2x^2 \equiv 3 \mod5$.

Since $2$ is a primitive root for $\mod5$, we have $2i\equiv3 \mod4$. But $\gcd(2,4)=2$ and $2$ does not divide $3$, hence there are no solutions.

Is this correct? Where would I factor in the $2$ from the original congruence? I suspect there is indeed a solution but my calculation does not take the $2$ from $2x^2 \equiv 3 \mod5$ into consideration. Any help will be greatly appreciated.

• Shouldn't it be $2i \equiv 3 \pmod{5}$? Commented Mar 25, 2014 at 21:27
• There are two solutions, $x\equiv \pm 2$. Commented Mar 25, 2014 at 21:31
• No, because when equating the exponents, we have to solve in mod(phi(n))=mod(phi(5))=mod(4). Commented Mar 25, 2014 at 21:31
• Note that there are only $5$ numbers modulo $5$, so finding the solutions directly is always easy. Commented Mar 25, 2014 at 21:32

An easier way than you're attempting: 3 is the multiplicative inverse of 2 modulo 5, so to solve $2x^2 \equiv 3\; (\textrm{mod}\; 5)$, we can multiply through by 3 and look for solutions to $x^2 \equiv 9 \equiv 4\; (\textrm{mod}\; 5)$, whereby we see $x = \pm2$ are the only solutions.
Or, $\ 2x^2\equiv -2\,\Rightarrow\, x^2\equiv -1\equiv 4\,\Rightarrow\,x\equiv \ldots$
If you rewrite the original equation you have $$2(y)\equiv 3 \mod 5 \implies y \equiv 4$$ The questions then becomes $$x^2 \equiv 4 \mod 5$$ which is true for $x\equiv \pm2$