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for each prime p there is k : A(k),A(k + 1) are divisible by p, trinomial

Source: 2010 Dutch IMO TST2 p5

January 10, 2020
number theorytrinomial

Problem Statement

The polynomial A(x)=x2+ax+bA(x) = x^2 + ax + b with integer coefficients has the following property: for each prime pp there is an integer kk such that A(k)A(k) and A(k+1)A(k + 1) are both divisible by pp. Proof that there is an integer mm such that A(m)=A(m+1)=0A(m) = A(m + 1) = 0.
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