Local Mathematical Competitions

We will prove the inequality $5^{n+1}\ge n^6$ holds for all positive integers $n$, except $n=3$ and $n=4$.

Indeed, this is readily verified for $n=1$, $n=2$ and $n=5$. When the thesis is true for $n\ge 5$, then $5^{n+2}=5\cdot 5^{n+1}\ge 5n^6$. It is then sufficient to show that $5n^6\ge (n+1{)}^6$, rewritten as $5\ge {\left(1+\frac{1}{n}\right)}^6$. Since ${\left(1+\frac{1}{n}\right)}^6<{\left(1+\frac{1}{4}\right)}^6=5\cdot \frac{5^5}{4^6}=5\cdot \frac{3125}{4096}<5$, everything is thus proven, by simple induction; moreover, the inequality becomes strict for $n>5$.

One thus gets $5^{a+1}\ge a^6$ and $5^{b+1}\ge b^6$, for all $a,b$ different from $3$ and $4$. Multiplying, first the two inequalities given in the problem statement, then those two just obtained in the above, we get $a^6{b}^6\ge 5^{b+1}{5}^{a+1}\ge b^6{a}^6$. But equality only holds here if $a=b=5$, which clearly checks. Otherwise, it needs $a=4$ or $a=3$ (or, symmetrically, $b=4$ or $b=3$). If $a=4$, then the inequalities given in the problem statement become $4^6>5^{b+1}$ and $b^6>5^5$, leading to $b\le 4$ and $b\ge 4$ respectively, hence $b=4$. If $a=3$, then $3^6>5^{b+1}$ and $b^6>5^4$, leading to $b\le 3$ and $b\ge 3$ respectively, hence $b=3$.

Two more pairs of solutions $a=b=4$ and $a=b=3$ have thus been found.