48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Suppose that a pair $(k,n)$ satisfies the condition of the problem. Since $7^k-3^n$ is even, $k^4+n^2$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^4+n^2\equiv 1+1=2(\mathrm{mod}4)$, while $7^k-3^n\equiv 7-3\equiv 0(\mathrm{mod}4)$, so $k^4+n^2$ cannot be divisible by $7^k-3^n$. Hence, both $k$ and $n$ must be even.

Write $k=2a$, $n=2b$. Then $7^k-3^n=7^{2a}-3^{2b}=\frac{7^a-3^b}{2}\cdot 2\left(7^a+3^b\right)$, and both factors are integers. So $2\left(7^a+3^b\right)\mid 7^k-3^n$ and $7^k-3^n\mid k^4+n^2=2\left(8a^4+2b^2\right)$, hence $$7^a+3^b\le 8a^4+2b^2.\text{\hspace{1em}(1)}$$ We prove by induction that $8a^4<7^a$ for $a\ge 4$, $2b^2<3^b$ for $b\ge 1$ and $2b^2+9\le 3^b$ for $b\ge 3$. In the initial cases $a=4$, $b=1$, $b=2$ and $b=3$ we have $8\cdot 4^4=2048<7^4=2401$, $2<3$, $2\cdot 2^2=8<3^2=9$ and $2\cdot 3^2+9=3^3=27$, respectively.

If $8a^4<7^a$ $(a\ge 4)$ and $2b^2+9\le 3^b$ $(b\ge 3)$, then $$\begin{array}{rl}8(a+1{)}^4& =8a^4{\left(\frac{a+1}{a}\right)}^4<7^a{\left(\frac{5}{4}\right)}^4=7^a\frac{625}{256}<7^{a+1}\text{ and }\\ 2(b+1{)}^2+9& <\left(2b^2+9\right){\left(\frac{b+1}{b}\right)}^2\le 3^b{\left(\frac{4}{3}\right)}^2=3^b\frac{16}{9}<3^{b+1},\end{array}$$ as desired.

For $a\ge 4$ we obtain $7^a+3^b>8a^4+2b^2$ and inequality (1) cannot hold. Hence $a\le 3$, and three cases are possible.

Case 1: $a=1$. Then $k=2$ and $8+2b^2\ge 7+3^b$, thus $2b^2+1\ge 3^b$. This is possible only if $b\le 2$. If $b=1$ then $n=2$ and $\frac{k^4+n^2}{7^k-3^n}=\frac{2^4+2^2}{7^2-3^2}=\frac{1}{2}$, which is not an integer. If $b=2$ then $n=4$ and $\frac{k^4+n^2}{7^k-3^n}=\frac{2^4+4^2}{7^2-3^4}=-1$, so $(k,n)=(2,4)$ is a solution.

Case 2: $a=2$. Then $k=4$ and $k^4+n^2=256+4b^2\ge |7^4-3^n|=|49-3^b|\cdot (49+3^b)$. The smallest value of the first factor is 22, attained at $b=3$, so $128+2b^2\ge 11(49+3^b)$, which is impossible since $3^b>2b^2$.

Case 3: $a=3$. Then $k=6$ and $k^4+n^2=1296+4b^2\ge |7^6-3^n|=|343-3^b|\cdot (343+3^b)$. Analogously, $|343-3^b|\ge 100$ and we have $324+b^2\ge 25(343+3^b)$, which is impossible again.

We find that there exists a unique solution $(k,n)=(2,4)$.