IMO 1J, Mock Exam 2

Let $p^a + a^4 = b^2$ for some positive integer $b$. Then we have
$$p^a = b^2 - a^4 = (b + a^2)(b - a^2).$$
Hence both $b + a^2$ and $b - a^2$ are powers of $p$.
Let $b - a^2 = p^x$ for some integer $x$. Then $b + a^2 = p^{a-x}$ and $a - x > x$. Therefore we have
$$2a^2 = (b + a^2) - (b - a^2) = p^{a-x} - p^x = p^x(p^{a-2x} - 1).$$
We shall consider two cases according to whether $p = 2$ or $p \neq 2$. We let $v_p(m)$ denote the $p$-adic valuation of $m$.

Case 1 ($p = 2$): In this case,
$$a^2 = 2^{x=1}(2^{a-2x} - 1) = 2^{2v_2(a)}(2^{a-2x} - 1),$$
where the second equality comes from $\text{gcd}(2, 2^{a-2x} - 1) = 1$. So, $2^{a-2x} - 1$ is a square.
If $v_2(a) > 0$, the $2^{a-2x}$ is also a square. So, $2^{a-2x} - 1 = 0$ and $a = 0$ which is a contradiction.
If $v_2(a) = 0$, then $x = 1$, and $a^2 = 2^{a-2} - 1$. If $a \ge 4$, the right hand side is congruent to 3 modulo 4, thus cannot be a square. It is easy to see that $a = 1, 2, 3$ do not satisfy this condition.
Therefore, we do not get any solutions in this case.

Case 2 ($p \neq 2$): In this case, we have $2v_p(a) = x$. Let $m = v_p(a)$. Then we have $a^2 = p^{2m} \cdot n^2$ for some integer $n \ge 1$. So, $2n^2 = p^{a-2x} - 1 = p^{p^m \cdot n-4m} - 1$.
We consider two subcases.

Subcase 2-1 ($p \ge 5$): By induction one can easily prove that $p^m \ge 5^m > 4m$ for all $m$. Then we have
$$2n^2 + 1 = p^{p^m \cdot n - 4m} > p^{p^m \cdot n - p^m} \ge 5^{5^{m \cdot (n-1)}} \ge 5^{n-1}$$
But by induction, one can easily prove that $5^{n-1} > 2n^2 + 1$ for all $n \ge 3$. Therefore, we conclude that $n = 1$ or $2$. If $n = 1$ or $2$, then $p = 3$, which is a contradiction. So there are no solutions in this subcase.

Subcase 2-2 ($p = 3$): Then we have $2n^2 + 1 = 3^{3m \cdot n - 4m}$. If $m \ge 2$, one can easily prove by induction that $3^m > 4m$. Then we have
$$2n^2 + 1 = 3^{3m \cdot n - 4m} > 3^{3m \cdot (n-1)} \ge 3^{9(n-1)}$$
By induction, one can easily prove that $3^{9(n-1)} > 2n^2 + 1$ for all $n \ge 2$. Therefore, we conclude that $n = 1$. Then we have $2 \cdot 1^2 + 1 = 3^{3m-4m}$ hence $3^m - 4m = 1$. The only solution of this equation is $m = 2$ in which case we have $a = 3^m \cdot n = 3^2 \cdot 1 = 9$.
If $m \le 1$, then there are two possible cases: $m = 0$ or $m = 1$.
If $m = 1$, then we have $2n^2 + 1 = 3^{3n-4}$. By induction, can see that $3^{3n-4} > 2n^2 + 1$ for all $n \ge 3$. By checking $n = 1, 2$, we only get $n = 2$ as a solution, this gives $a = 3^m \cdot n = 3^1 \cdot 2 = 6$.
If $m = 0$, then we have $2n^2 + 1 = 3^n$. By induction, can see that $3^n > 2n^2 + 1$ for $n \ge 3$. Checking $n = 1, 2$, we find the solution $a = 3^0 \cdot 1 = 1$ and $3^0 \cdot 2 = 2$. Therefore, $(a, p) = (1, 3), (2, 3), (6, 3), (9, 3)$ are all possible solutions.