Competition problem — Algebra — Maths Olympiad Prep

Algebra Difficulty 4.8 AIME Prove it Romania

Find all integers $n \ge 0$ for which there exist integers $a$ and $b$ such that $a + 2^b = n^{2022}$ and $a^2 + 4^b = n^{2023}$ .

Solution

The only number is $n = 1$ .
To this end, notice that $2(a^2 + 4^b) \ge (a + 2^b)^2$ to conclude that $2n^{2023} \ge n^{4044}$ , hence $n = 0$ or $n = 1$ . The case $n = 0$ leads to no solution, while $n = 1$ holds for $a = 0$ and $b = 0$ .

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