<s>[INST] How to use prove this $p^4\equiv p\pmod {13}$ 
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$

A friend of mine suggested that I might be able to use the results problem.
 [/INST] We will have modulo $13$ all the time. Notice that $$N:= n^4+n^3+2n^2-4n+3 \equiv (n^2-6n-4)^2 \equiv (n-3)^4$$
Clearly $13\mid N\iff 13\mid n-3$, so we assume $p\ne 13$ and thus $13$ does not divide $n-3$. Say prime divisor $p$ of $N$ is good if $p^4{\equiv}p$. 
Suppose the statment is not true, so there exists a prime $q$ which is not good. 
Also we can assume that there is no good prime divisor of $N$: if $p\ne 13$ is good than we can divide $N$ by $p\equiv p^4$ modulo $13$ and then observe $N' = N/p^4$ modulo $13$.  Also we can reduce $N$ with all divisors of $N$ of form $d^4$. So if we make now a prime factorisation for $N$ modulo $13$ we have $$N\equiv  2^a4^b5^c6^d7^e8^f10^g11^h12^i$$
(clearly there is no $0,1,3$ and $9$) where all the exponents are nonnegative and less than $4$ and at least one is positive. Of course, we can reduce this even more:
$$N\equiv  2^{a+2b+d+3f+g+2i} 3^{d+i} 5^{c+g}7^e11^h$$
or $$N\equiv  2^x 3^y 5^z(-6)^e(-2)^h$$
where or exponents are nonegative and less than $4.$ Again we can assume if $y=0$ (else we divide $N$ by $3$ modulo $13$). So we can write:$$ N\equiv  (-1)^{3z+e+h} 2^{x+3z+e+h} = (-1)^t2^{x+t}$$
where $t= 3z+e+h$. So $$N\equiv   -1\pm2,\pm4,\pm8$$
Now this can not be of the form $(...)^4$ and we have a contradiction.
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