Suppose \(f \equiv g \bmod p\) and \(f_1 \equiv g_1 \bmod p\).
Prove that \(f + f_1 \equiv g + g_1 \bmod p\).
Prove that \(ff_1 \equiv gg_1 \bmod p\).
Solution
Assume \[ p\mid(f-g),\qquad p\mid(f_1-g_1). \] For sums, \[ (f+f_1)-(g+g_1)=(f-g)+(f_1-g_1), \] which is divisible by \(p\). Hence \[ f+f_1\equiv g+g_1\pmod p. \]
For products, \[ ff_1-gg_1=f(f_1-g_1)+g_1(f-g). \] Both terms on the right are divisible by \(p\), so \[ ff_1\equiv gg_1\pmod p. \]