by Anil
Last Updated April 28, 2017 15:20 PM

we started recently into elementary number theory and I´m stuck at solving the following quadratic modular equation.

x^2 +15 ≡ 0 (mod 8)

According to WolframAlpha the solution is x=1+2n (n∈Z)

Can anyone explain me how to get this solution?

Best regards

One can proceed naively: Reducing modulo $2$ gives the equation $$x^2 + 1 \equiv 0 \pmod 2 .$$ So, $x \equiv 1 \pmod 2$, that is, any solution must have the form $x = 2 n + 1$. Substituting into our original equation gives $$(2 n + 1)^2 + 15 = 0 \pmod 8 ,$$ and expanding gives $$4 n^2 + 4 n = 0 \pmod 8 ,$$ which is equivalent to $$n^2 + n = 0 \pmod 2.$$ This equation is true for every $n$, so all quantities $x = 2 n + 1$ are solutions to the original equation.

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