parmenides51 30.09.2018 20:14 Prove that for all positive integers $n$, $169$ divides $21n^2 + 89n + 44$ if $13$ divides $n^2 + 3n + 51$.
kevinmathz 30.09.2018 20:17 perhaps factor? Also, note you can add 169 * n to the left, and 13 * m to the right, and it'll be the same proof(assuming n, m integers)
Illuzion 30.09.2018 21:00 With inspection we get that the given condition is satisfied when $n=13k+5$ and so $21n^2+89n+44=21*169k^2+21*130k+21*25+89*13k+89*5+44=169k^2+169*23k+169*6$ and the last is clearly divisible by 169.