# How to solve this quadratic congruent equation by inspection

I found a systematic way (c.f. How to solve this quadratic congruence equation) to solve all congruent equations of the form of $$ax^2+bx+c=0\pmod{p}$$, or to determine that they have no solution.

But I wonder if there is some easy way to find solutions of simple quadratic congruent equations by inspection, for example, for this equation $$x^2+x+47=0\pmod{7}.$$

My textbook gave the solutions $$x\equiv 5\pmod{7}$$ and $$x\equiv 1\pmod{7}$$ directly, without any proof. So I think there must be some easier way to inspect the solutions.

Any help will be appreciated.

• Note: $x^2+x+47\equiv x^2-6x+5\pmod 7$ – J. W. Tanner May 10 '19 at 14:23
• Also you can note that $x(x+1)=-5=2$ – Piquito May 10 '19 at 14:35
• @J.W.Tanner What an excellent observation! Thanks Tanner. – Sam Wong May 10 '19 at 20:54
• @Piquito Thanks for your observation too. But I think Tanner's observation is more straightforward :) – Sam Wong May 10 '19 at 20:55

The good old quadratic equation works just fine if $$p\neq2$$. If $$a\not\equiv 0\pmod{p}$$ and $$b^2-4ac$$ is a quadratic residue mod $$p$$, then the solutions to the quadratic congruence $$ax^2+bx+c\equiv0\pmod{p},$$ are precisely $$-\frac{b\pm\sqrt{b^2-4ac}}{2a}.$$

In this particular case we have $$p=7$$ and $$a\equiv b\equiv1\pmod{7}$$ and $$c\equiv47\equiv5\pmod{7}$$. Then $$b^2-4ac\equiv2\equiv3^2\pmod{7},$$ so the congruence has the two solutions $$-\frac{1+3}{2}=5\qquad\text{ and }\qquad-\frac{1-3}{2}=1.$$

On the other hand, if you want to solve it purely by inspection, note that there are only $$7$$ possible solutions to check. Clearly $$x=0$$ is not a solution, and plugging in $$x=1$$ yields the first solution. The product of the solutions is congruent to $$47\equiv5\pmod{7}$$, so the other solution is $$x=5$$.

• Thanks Servaes. But I still have problems in understanding your last sentence. Why is that the product of the solutions is congruent to $47\equiv 5\, ($mod$\, 7)$ ? And why if it is so, the other solution will be $x=5$? Thanks for your elaboration. – Sam Wong May 10 '19 at 20:54
• @SamWong: the product of the roots of $x^2+bx+c=0$ is $c$, so $5$ in this case, and the solution of $1\times x\equiv5$ is $x\equiv5$ – J. W. Tanner May 10 '19 at 21:10
• @J.W.Tanner Oh I see. Thanks! – Sam Wong May 10 '19 at 22:41

Servaes gave a good general method for solving quadratic equations modulo a prime.

I will elaborate on my comment about how in this particular case the answer could be found by inspection.

Adding or subtracting a multiple of $$p$$ to or from any of the coefficients $$a,b,c$$

does not materially change the equation $$ax^2+bx+c\equiv0\mod p.$$

In fact, $$x^2+x+47$$ begs to be reduced modulo $$7$$ to $$x^2+x+5.$$

$$x^2+x+5$$ doesn't have rational roots, but you could add or subtract any multiple of $$p$$ to the $$x$$-coefficient $$1$$,

yielding polynomials such as $$x^2+8x+5, x^2-6x+5,$$ etc.,

and $$x^2-6x+5$$ is easily recognized as $$(x-1)(x-5)$$.