I assume that x₁, x₂, x₃, x₄, and x₅ are integers, otherwise we cannot have finite solution

Also, are there any restrictions on x₄ and x₅, such as x₄ >= 0, x₅ >= 0 ?

Otherwise, there are infinitely many solutions: all you need to do is set x₁, x₂ and x₃ to any valid number, then take any value of x₄ and find the value of x₅ that will make equation true

For example x₁ = 3, x₂ = 2, x₃ = 20

Then we can have x₄ + x₅ = 21 − 20 − 3 − 2 = −4

and there are infinitely many pairs of integers whose sum is −4

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Really nice solution by iceman. Took me a moment to figure out what he was doing.

Anyway, we can make product simpler. Since x₂ ≥ 1, then x₃ ≤ 20.

Also, since x₁ + x₂ + x₃ ≥ 16, x₄ and x₅ ≤ 5. So we get:

(∑x^k, k = 0 to 3) * (∑x^k, k = 1 to 3) * (∑x^k, k = 15 to 20) * (∑x^k, k = 0 to 5)²

= x^36 + 5x^35 + ... + 157x^22 + 106x^21 + 64x^20 + ... + 5x^17 + x^16

Solution remains the same: coefficient of x^21 = 106