d is not divisible by 5. For some integer n, a n3 + b n2 + c n + d is divisible by 5. Show that for some integer m, a + b m + c m2 + d m3 is divisible by 5.
Solution
Evidently we need only consider n mod 5. Since d ≠ 0 mod 5, we must have n ≠ 0 mod 5. If n = 1 mod 5, then a+bn+cn2+dn3 = an3+bn2+cn+d mod 5, so can take m = 1. If n = 4 mod 5, then an3+bn2+cn+d = 4a+b+4c+d mod 5. If 4a+b+4c+d = 0 mod 5, then a+4b+c+4d = 0 mod 5, and since a+4b+42c+43d = a+4b+c+4d, we can take m = 4.
If n = 2 mod 5, then an3+bn2+cn+d = 3a+4b+2c+d mod 5. So if this is 0 mod 5, then so is 2(3a+4b+2c+d) = a+3b+4c+2d = a+3b+32c+33d mod 5, so we can take m = 3.
If n = 3 mod 5, then an3+bn2+cn+d = 2a+4b+3c+d mod 5. So if this is 0 mod 5, then so is 3(2a+4b+3c+d) = a+2b+4c+3d = a+2b+22c+23d mod 5, so we can take m = 2.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03