If ac, bc + ad, bd = 0 (mod n) show that bc, ad = 0 (mod n).
Solution
We have (bc - ad)2 = (bc + ad)2 - 4ac·bd. So ( (bc-ad)/n)2 = ( (bc+ad)/n)2 - 4(ac/n)(bd/n), which is an integer. Hence (bc - ad)/n is an integer. Moreover ( (bc+ad)/n)2 - ( (bc-ad)/n)2 is an even integer, so (bc+ad)/n and (bc-ad)/n have the same parity. Hence bc/n = half their sum, and ad/n = half their difference, are both integers.
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03