2008 OIM Problems/Problem 4

Problem

Prove that there are no positive integers $x$ and $y$ such that

\[x^{2008}+2008!=21^y\]

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

Clearly $21|2008!$, so $21|x^{2008}$ as well. Then $21|x$, so let $x=21a$ for some positive $a$. Our equation would become: \[21^{2008}a^{2008}+2008!=21^y\] But notice that \[21^y=21^{2008}a^{2008}+2008!>21^{2008}a^{2008}\ge 21^{2008}\] implying that $y>2008$. This will be used later.

Next, we note that by Legendre's Formula, there are $331$ powers of $7$ and $1000$ powers of $3$ in the prime representation of $2008!$. Then we can let $2008!=3^{1000}7^{331}b$ for some positive integer $b$ that is not divisible by $3$ or $7$. Substituting: \[21^{2008}a^{2008}+3^{1000}7^{331}b=21^y\] \[\Rightarrow 21^{2008}a^{2008}+3^{669}21^{331}b=21^y\] \[\Rightarrow 21^{1677}a^{2008}+3^{669}b=21^{y-331}\] \[\Rightarrow 3^{669}b=21^{y-331}-21^{1677}a^{2008}\] Since $y>2008>331$, the right-hand side is divisible by $7$. However, the left-hand side is not, which is not possible; thus there are no positive integer solutions to the equation.

~ eevee9406

See also

OIM Problems and Solutions