, 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 $m^2 2^a \equiv 2 \pmod{3}$ 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 $3^b \equiv 1 \pmod{11}$ 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 $a \ge 2$.