Tìm các số tự nhiên m,n sao cho n^2 + 25 = 27(m+n) - 3^m
$n^2 + 25 = 27(m+n)-3^{m}$
$m\geq 2,VP= 27(m+n)-3^{m}\equiv 0 (mod9)$
$25\equiv 7 (mod9)$
$n^{2}\equiv 0,1,4,7 (mod9)$
$=> VT\equiv 7,8,2,5 (mod9)$
=> Vô lí
$TH. 0\leq m<2 => m=0=>n=26,n=1$
m=1...