Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$

Question

Find all natural number(s) $n$ such that $55\mid n^2 + 3n + 1$

I see that $n^2+3n+1 =n^2-2n+1+5n\equiv n^2-2n+1=(n-1)^2 \pmod{5}$
I also see that $n^2+3n+1=n^2+3n-10+11=(n-2)(n+5)+11\equiv(n-2)(n+5) \pmod{11}$
After that what?

Answer 1

from what you deduced we have $n\equiv 1 \bmod 5$ and $n\equiv 2$ or $n\equiv 6 \bmod 11$.

This tells us $n\equiv 46$ or $n\equiv 6 \bmod 55$.

Answer 2

There are infinitely many solutions, in particular you have the identity $$(55t-119)^2+3(55t-119)+1=55(55t^2-235t+251)$$which shows solutions $x=55t-119$ for each integer $t$ you choose. There are maybe other solutions out of this infinity of them.