Competition problem — Number theory

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Number theory Difficulty 5.7 AIME, harder Prove it

3 Prove: There are infinitely many odd numbers $m$ , such that $8^{m}+9 m^{2}$ is a composite number.

Solution

3. Take $m=9 k^{3}(k=1,3, \cdots)$ , then $8^{m}+9 m^{2}=\left(2^{m}\right)^{3}+\left(9 k^{2}\right)^{3}$ . It is easy to see that it has a proper divisor $2^{m}+9 k^{2}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.