Team Selection Test for IMO 2024

Answer: $(a,b)=(1,1)$ and $(1,2)$. Assume that $$\frac{10^a-3^b+1}{2^a}=m^2$$ holds for some integer $m$. Let us rewrite the equation as $$3^b=10^a-m^2{2}^a+1.$$ If $a=1$ then clearly the only solutions are $(a,b)=(1,1)$ and $(1,2)$. If $a\ge 2$ then $a$ is even. Then, since $3$ divides $10^a-m^2{2}^a+1$, we get $m^2{2}^a\equiv 2(\mathrm{mod}3)$ which means that $a$ is odd. If $a=3$ then by using modulo $4$ we get that $b$ is even. Since $10^6+1=3^6+m^2{2}^a=c^2+2d^2$, we have no integer solutions because $-2$ is not a quadratic residue modulo $101$ and $101\nmid 3^{b/2}$. Therefore, $a\ge 5$. In that case, by using modulo $16$ we get that $4\mid b$ and hence $5\mid 3^b-1$. Then $5\mid m$ and hence $25\mid 3^b-1$. Let $b=4k$, then $25\mid 81^k-1$ and from the LTE lemma, $$v_5(81^k-1)=v_5(80)+v_5(k)\ge 2$$ and hence $5\mid k$, which implies $5\mid b$. Hence $3^b\equiv 1(\mathrm{mod}11)$ and $11\mid 10^a-m^2{2}^a=x^2-2y^2$ but $2$ is not a quadratic residue modulo $11$ so we must have $11\mid 10^{a/2}$, which is a contradiction. Thus, there is no solution for $a\ge 2$.