Problem:
Let p be the least prime number for which there exists an integer n such that n4+1 is divisible by p2. Find the least positive integer m such that m4+1 is divisible by p2.
Solution:
First, note that pξ =2, because n4+1 is not divisible by 4 for any integer n. Moreover, if n4+1β‘0(modp), then n4β‘β1(modp) and n8β‘1(modp), so n has order 8, implying that 8β£(pβ1). The least prime that satisfies this condition is p=17. So, consider whether there is an integer n such that 172 divides n4+1.
Let m=17k+r, with 0β€rβ€16. Then
m4+1=(17k+r)4+1=172(172k4+4β
17k3r+6k2r2)+4β
17kr3+r4+1β
is divisible by 172 if and only if 4kβ
r3+17r4+1β is divisible by 17. The solutions to r4+1β‘0(mod17) are r=2,8,9, and 15. For each r, the least values of k and m are given by the following ordered triples
(r,k,m)=(2,9,155),(8,6,110),(9,10,179),(15,7,134).β
Thus the least possible prime p is 17, and the least positive integer m for which 172 divides m4+1 is 110β.
The problems on this page are the property of the MAA's American Mathematics Competitions