ALG-B1-M04-P018 Many Trinomials with Integer Roots

First prove: if \(ax^2+bx+c\) has integer roots, then \(a\mid b\) and \(a\mid c\).

By Vieta for integer roots \(r,s\): \(-b/a=r+s\), \(c/a=rs\). Hence \(a\mid b\) and \(a\mid c\).

Suppose the numbers used are \(k+1,\ldots,k+3n\). Order the leading coefficients: \(a_1<\cdots

For the trinomial with leading coefficient \(a_n\), the coefficients \(b,c\) must be positive multiples of \(a_n\), so one of them is at least \(3a_n\). But every coefficient is at most \(k+3n\). Thus \(3(k+n)\le k+3n\), hence \(k=0\). Then \(a_i=i\).

If \(i\) is even, all three coefficients in that trinomial are even. If \(i\) is odd, at least one coefficient in the triple must be even; otherwise the trinomial has odd value at every integer \(x\) and cannot have an integer root. Thus the number of even coefficients is at least \(n+2\lfloor n/2\rfloor\), which for \(n>10\) exceeds the number of even integers among \(1,\ldots,3n\). Contradiction.

The regional idea is scaled down, preserving the divisibility lemma and parity count.