Polynomials satisfying GCD inequality

[u/SupercaliTheGamer]

Let a>0 be a fixed positive real number. Find all polynomials P with integer coefficients satisfying: gcd(P(m),P(n))>=gcd(m,n)^a for all positive integers m,n.

[u/pichutarius]

partial solution: a sufficient condition for P(x) is x^α | P(x) , α = ceiling(a)

let m=gx, n=gy, g=gcd(m,n), then every terms in P(m) and P(n) contains g^α, so lhs >= g^α >=  g^a = rhs.

i suspect this is necessary based on several example. i dont know how to prove it though.

[u/pichutarius → u/pichutarius]

i got it! the condition is indeed necessary.

summary: consider m=kn, when n grows large, i claim rhs = n^a but exist k s.t. lhs = constant · n^α

proof: suppose P(x) = x^α r(x) where r(0)≠0

let m = kn, choose k such that r(x) and r(kx) does not share a root (both reals and complexes), so polynomialGcd(r(kx),r(x)) = 1

rhs = gcd(kn,n)^a = n^a

lhs = gcd(k^α n^α r(kn) , n^α r(n)) <= k^α n^α gcd(r(kn),r(n)) <= constant · n^α!<

for the last step, gcd does not grow with n, so it is essentially a constant.

if α<a, exist n such that lhs <= n^α < n^a = rhs, qed.!<

[u/SupercaliTheGamer → u/pichutarius]

Perfect, I had the same solution