Two polynomial level sets at 54 and 2013: the factorization bound

LobaLootedIt·Aug 9, 2026(edited)

The difference 1959 factors as

3\cdot653

. I can bound each level set separately, but I'm not convinced that this proves the claimed sharp bound on their product. Problem: A polynomial

P

with integer coefficients satisfies

P(x_1) = P(x_2) = \dots = P(x_k) = 54

and

P(y_1) = P(y_2) = \dots = P(y_n) = 2013

for distinct integers

x_1, \dots, x_k; y_1, \dots, y_n

. Determine the maximal value of

kn

. Source: Baltic Way 2013 · MNET-019N

Open MNET-019N in the question bank →

3 replies

tarnished_ish·Aug 10, 2026(edited)

For a fixed integer x with P(x)=54, factor

P(t)-2013

by all its distinct integer roots y_j. Evaluating at x shows

\prod_j(x-y_j)

divides −1959. The factors are distinct nonzero integers, so at most two have absolute value 1, and at most two more can consume the prime factors 3 and 653. Thus n≤4; similarly k≤4.

LobaLootedIt·Aug 10, 2026(edited)

That's stronger than counting the eight signed divisors separately. The product restriction is essential. But it still leaves 3×3, 3×4, and 4×4 to exclude.

MissingNo404·Aug 10, 2026

For the lower bound,

P(t)=653t^2(t^2-4)+2013

has value 2013 at t=0,±2, and value 54 at t=±1, giving kn=6. The upper bound still needs an argument linking the two level sets; separate size bounds alone do not finish it.

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