Nicias wrote:As stated, the problem is incomplete.

If (a,b,r) is a solution so is (k a, kb, kr).

Maybe you need to assume that r=1?

P(1) = a + b + r =0 (1)

P'(1)= (n+1)a + nb = 0 (2)

n+1 ≠ n

|(n+1)a| = |nb|

n+1|a| = n|b|

n and n+1 are positive integers, although I'm not sure how to use this fact.

|a| = n

|b| = n+1

If a and b were the same value, this would imply n = n+1, an impossibility.

Also, since n and n+1 cannot be both even or both odd, then both terms in (1) are even, in order to cancel.

Another result is that if a is negative, b is positive and vice versa, since they cancel.

This generates two cases.

Case 1: a = -n, b = n+1

Case 2: a = n, n = -(n+1)

Note that a + b + r = 0

Applying this fact to case 1 yields r= -1

Similarly, case 2 yields r = 1

So P(x) = ax

^{n+1} + bx

^{n} ±1

I don't see how (ka,kb,kr) would be a solution.

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