Argentine National Olympiad 2015

If $a=1$ or $b=1$ we obtain the solutions $(1,2^n-1)$ and $(2^n-1,1)$, with $n>1$.

Let $a,b\ge 2$ and $a<b$. Note that then $a+b<ab+1$ due to the identity $(ab+1)-(a+b)=(a-1)(b-1)$. Let $a+b=2^n$, $n\ge 2$; in fact then $n\ge 3$ as $n=2$ forces $a=b=2$. Then $a=2^{n-1}-c$, $b=2^{n-1}+c$ with $1\le c<2^{n-1}$. We have $$2^n=a+b<ab+1=2^{2(n-1)}-c^2+1\le 2^{2(n-1)},$$ which implies $ab+1=2^k$ with $n+1\le k\le 2(n-1)$. Write $$2^k=2^{2(n-1)}-c^2+1,$$ i.e., $ab+1=2^k$, in the form $$(c-1)(c+1)=2^{2(n-1)}-2^k.$$ Then $(c-1)(c+1)$ is divisible by $2^k$ since $k\le 2(n-1)$. On the other hand $n+1\le k$, so $(c-1)(c+1)$ is divisible by $2^{n+1}$. Clearly $c-1$ and $c+1$ are consecutive even numbers. Since one of them is divisible by $2$ but not by $4$, the other one is divisible by $2^n$. Now the inequalities $1\le c<2^{n-1}$ show that the later is possible only if $c-1=0$, i.e., $c=1$. Hence $a=2^{n-1}-1$, $b=2^{n-1}+1$ with $n\ge 3$. This is clearly an admissible pair. Hence the solutions are $(1,2^n-1)$, $(2^n-1,1)$ and $(2^n-1,2^n+1)$, $(2^n+1,2^n-1)$ with $n>1$.