5 is prime. A = {p - n2 where n2 < p}. Show that we can find integers a and b in A such that a > 1 and a divides b.
Solution
Take m2 to be the largest square less than p. Let p - m2 = k. Then p - (m - k)2 = (m2 + k) - (m2 - 2km + k2) = k(2m - k + 1). So p - m2 divides p - |m - k|2 (*).
Note that we cannot have k = m because then m would divide p. We cannot have (m - k) = m because then p = m2. We cannot have (k - m) = m, because then p = m2 + 2m = m(m+2). So (*) provides suitable a and b unless k = 1.
If k = 1, them m must be even (otherwise p = m2 + 1 would be even), hence p - (m-1)2 = 2m divides p - 1 = m2.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002