Start by rewriting the given equation to make a the subject.

2^px a^p/(p+2) = 2² x 5

a^p/(p+2) = 2^{2 - p} x 5

a^p = (2^{2 - p} x 5) ^{p + 2} = 2^{(2 – p)(p + 2)} x 5^{(p + 2)}

a = (2^{(2 – p)(p + 2)} x 5^{(p + 2)} )^1/p = 2^{2/p - 1} x 5^{(1 + 2/p)}

For a to be an integer we require both 2/p - 1 and (1 + 2/p) to be non-negative integers.

To ensure this for 2/p - 1, p = 1 or 2.

and for (1 + 2/p), p = 1, 2 or – 2.

Therefore we can use p = 1 or 2.

If p = 1, then a = 2³ x 5³ = 10000.

If p = 2, then a = 2⁰x 5² = 25.

So there are two solutions:

p = 1 and a = 1000

p = 2 and a = 25

Challenge 289 : Identify the Integers