let k, m, n, m1, n1 be Integer; :: thesis: ( k divides m & k divides n implies k divides (m * m1) + (n * n1) )
assume ( k divides m & k divides n ) ; :: thesis: k divides (m * m1) + (n * n1)
then ( k divides m * m1 & k divides n * n1 ) by INT_2:2;
hence k divides (m * m1) + (n * n1) by Th4; :: thesis: verum