let $p$and $q=p+2$ be twin primes. consider the diophantine equation $(+)$ given by $n!+pq^2=(mp)^2$ $m\geq1$, $n\geq1$ i. if $m=p$,find the value of $p$. ii. how many solution quadruple $(p,q,m,n)$ does $(+)$ have ?
Problem
Source: Nigerian mthematics olympiad round 3 problem 4
Tags: number theory, Diophantine equation
starchan
21.02.2020 06:11
Hint for $i$ The equation reduces to $p^4 - p(p + 2)^2 = n!$ this comes from the fact that $m = p , q= p+2$ This implies that $p|n!$ As $p$ is prime this implies that $p \leq n$. Also note that from $p^4 - p(p + 2)^2 = n!$ we get $p^4 \equiv n! mod q$ As $p = q-2$ this gives $q|n! - 16$ Now clearly $q$ is n odd prime since twin primes are known to be odd So $q$ does not divide $n!$ which gives $q > n$ So we have $p + 1 \geq n \geq p$ which gives $n = p$ or $p +1$
kaede
21.02.2020 15:21
(ii)
Assume that $p\neq 3$.
Then $p\equiv 2\bmod 3$ and $q\equiv 1\bmod 3$.
So we have $ $$n!+2\equiv m^{2}\bmod 3$, which implies $n\leq 2$.
However, if $n\leq 2$, then $p\nmid n!$ , impossible.
So we must have $p=3$ and $q=5$.
Since $p\mid n!$ and $p^{2} \nmid n!$ , we must have $3\leq n< 6$.
If $n=3$, then $6+75=9m^{2}$, which implies $m=3$.
If $n=4$, then $24+75=9m^{2}$, which implies $m^{2} =11$, impossible.
If $n=5$, then $120+75=9m^{2}$, which implies $3m^{2} =65$, impossible.
So the only solution to the equation is $( p,q,m,n) =( 3,5,3,3)$.
Hence the number of solutions to the equation is $\boxed{1}$.