surprisingly not trivial for a P2

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Source: Monthly Contest KTOM March 2025 P2 Essay

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iStud

#1 h Yesterday at 3:52 AM

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Find all natural numbers $(m,n)$ such that
$2^{n!}+1\mid 2^{m!}+19$
Hint

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#2 h Yesterday at 4:36 AM

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First if $n>m$ , then if $m=1$ then $2^{n!}+1 \mid 21$ and there's no solution clearly for $n \ge 2$ , now if $m=2$ then $2^{n!}+1 \mid 23$ then there's also no sol by size for $n \ge 3$ , now for $m \ge 3$ notice that $2^{n!} \ge 16 \cdot 2^{m!}$ which then by size shows there's no sol as well.
If $m=n$ then $2^{n!}+1 \mid 18$ which only has sol when $n=1$ so $(m,n)=(1,1)$ works.
Now if $m>n$ then $n! \mid m!$ so notice that $2^{n!}+1 \mid 2^{2 \cdot m!}-361$ but also remember that $2^{n!}+1 \mid 2^{2n!}-1 \mid 2^{2m!}-1$ then $2^{n!}+1 \mid 360$ and here you have $n \le 3$ , clearly $2^6+1=65$ so it doesnt work, now check that $n=1$ and $n=2$ both work, if $n=1$ then $3 \mid 2^{m!}+19$ which means $3 \mid 2^{m!}+1$ and so if $m!$ is even it fails so it forces $m=1$ , contradiction. Now if $n=2$ then $5 \mid 2^{m!}-1$ , which holds true if $4 \mid m!$ and that is true for all $m \ge 4$ , for $m=3$ its $5 \mid 63$ which is false, therefore all pairs are $(m,n)=(1,1), (k,2)$ for $k \ge 4$ thus we are done :cool:

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