BMO 2018 Problem 4 - Solution - 10 May 2018 - MATHEMATICS COMPETITIONS | НАТПРЕВАРИ ПО МАТЕМАТИКА

Problem 4

Find all primes and such that $3p^{q-1}+1$ divides 11^p+17^p .

Solution

We will first look at the case $q \neq 2$ .

Then, at least an odd prime divisor $3p^{q-1}+1$ (There is always such , as if $p \neq 2$ , The $3p^{q-1}+1$ They are $\equiv 4 \pmod 8$ , and after $3p^{q-1}+1>4$ , will necessarily have a prime (unnecessary) divisor. If now p=2 , The $3 \cdot 2^{q-1}+1$ is unnecessary, and obviously has a prime superfluous divider).

Therefore, $r \mid 3p^{q-1}+1 \mid 11^p+17^p \Rightarrow (\dfrac{11}{17})^{2p} \equiv 1 \pmod r$ .

Let's order it $\dfrac{11}{17} \pmod r$ be it .

Then, by its definition $\rm ord$ , They are $x \mid 2p, \, x \mid r-1$ .

Therefore, $x \in \{1,2,p,2p \}$ .

If $x \in \{1,p\}$ , then if x=1 , They are $(\dfrac{11}{17})^1 \equiv 1 \pmod r \Rightarrow r \mid 17-11=6 \Rightarrow r \in \{2,3\}$ , inappropriate, since the is unnecessary, and $r \mid 3p^{q-1}+1 \equiv 1 \pmod 3$ . If you too x=p , we also have an contradiction .

Therefore, x=2 or x=2p .

If x=2 , then $(\dfrac{11}{17})^2 \equiv 1 \pmod r \Rightarrow r \mid 168 \Rightarrow r=7$ , since $r \neq 2,3$ .

This means that $3p^{q-1}+1=4^m7^n$ (after $3p^{q-1}+1 =3p^{2\ell}+1=3(p^2)^\ell +1 \equiv 0 \pmod 4$ ). From LTE, it is $7^2 \mid \mid 11^n+17^n$ , therefore $n \leqslant 2$ , and the like $m \leqslant 1$ . By taking the cases, we have the solution (p,q)=(3,3) .

If x=2p , the $r \equiv 1 \pmod {2p}$ , and so $3p^{q-1}+1=4^m7^ns$ , with $(s,14)=1, \, s \equiv 1 \pmod p$ . Like before, $m \leqslant 2, \, n \leqslant 1$ , and thus getting $\pmod p$ , we have got $3,27 \equiv 0 \pmod p \Rightarrow (p,q)=(3,3)$ .

Finally, we look at the case q=2 . Then, $3p+1 \mid 17^p+11^p$ . If the 3p+1 has an odd prime divisor, albeit , we apply the procedure of the previous solution. If not, then 3p+1=2^d , and after $(-1)^d \equiv 2^d=3p+1 \equiv 1 \pmod 3 \Rightarrow d=2t$ .

So, (2^t-1)(2^t+1)=3p . After (2^t-1,2^t+1)=1 , They are 2^t-1=1,2^t+1=3p or 2^t-1=3, 2^t+1=p . Both cases easy to prove . (the first one gives $t=1 \Rightarrow p=1$ ,contradiction, and the second p=5 , which does not meet the requirement ).

Ultimately, the only solution is $\boxed{(p,q)=(3,3)}$ .