One of the factors of (8 2k + 5 2k ), where k is an odd number, is:
89
Rewrite the exponents with a common power: \(8^{2k}=(8^2)^k=64^k\) and \(5^{2k}=(5^2)^k=25^k\). The expression becomes \(64^k+25^k\).
For any odd positive integer \(n\), \(a^n+b^n\) is divisible by \(a+b\) (substituting \(a=-b\) makes the polynomial zero, so \((a+b)\) is a factor).
Since \(k\) is odd, \(64^k+25^k\) is divisible by \(64+25=89\). Therefore 89 is a factor.
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