Slovenija 2008

Assume that $n^2-5^{4m+3}>0$. If $n^2-5^{4m+3}=1$, then $5^{4m+3}=n^2-1=(n-1)(n+1)$. Since $n$ is obviously even the numbers $n-1$ and $n+1$ must be coprime, which implies $n-1=1$ and $n+1=5^{4m+3}$. This is not possible. If $n$ is divisible by $5$, then $n^2-5^{4m+3}\ge 5$, if $n$ is of the form $n=5k\pm 1$, then

$n^2-5^{4m+3}\equiv 1(\mathrm{mod}5)$, and if $n=5k\pm 2$, then $n^2-5^{4m+3}\equiv 4(\mathrm{mod}5)$. Since none of the remainders are equal to $2$ or $3$ and $n^2-5^{4m+3}>1$, we conclude that $n^2-5^{4m+3}\ge 4$.

Now, assume $5^{4m+3}-n^2>0$. This number cannot be equal to either $2$ or $3$. We can see this by considering the remainders modulo $5$, because $5^{4m+3}-n^2$ is either divisible by $5$ (and hence greater than or equal to $5$), or $n$ has the form $n=5k\pm 1$ or $n=5k\pm 2$. So, either $5^{4m+3}-n^2\equiv -1\equiv 4(\mathrm{mod}5)$ or $5^{4m+3}-n^2\equiv -4\equiv 1(\mathrm{mod}5)$.

Assume there exist $m$ and $n$ such that $5^{4m+3}-n^2=1$. We can rewrite the equation as $$\begin{array}{rl}{n}^2& =5^{4m+3}-1=(5-1)(5^{4m+2}+5^{4m+1}+\cdots +5+1)=\\ & =4(5^{4m+2}+5^{4m+1}+\cdots +5+1).\end{array}$$ This implies that $5^{4m+2}+5^{4m+1}+\cdots +5+1$ is a perfect square. If it is equal to $l^2$, then $(l-1)(l+1)=5(5^{4m+1}+5^4+\cdots +5+1)$. Let us consider the expression $5^{4m+1}+5^4+\cdots +5+1$ modulo $4$. We have $$5^{4m+1}+5^4+\cdots +5+1\equiv 1+1+\cdots +1+1\equiv (4m+2)\equiv 2(\mathrm{mod}4),$$ since there are $4m+2$ summands in the sum. This sum is therefore divisible by $2$ but not by $4$. So, $(l-1)(l+1)$ is even. On the other hand, $l-1$ and $l+1$ have the same parity, so $(l-1)(l+1)$ is divisible by $4$, a contradiction. The numbers $m$ and $n$ such that $5^{4m+3}-n^2=1$ do not exist.

We have shown that $n^2-5^{4m+3}$ and $5^{4m+3}-n^2$ cannot be equal to $0$, $1$, $2$ or $3$, so the smallest possible value of $|5^{4m+3}-n^2|$ is $4$ and this value is attained when $n=11$ and $m=0$, for example.