Show that no square number has a decimal ending in 79. More generally, find all possible two-digit endings for squares.
Let any digit number ending at 79 be represented as
$$a_nx^n+.....+7x+9$$
Plug in 10
$$a_n10^n+.....+79$$
square the expression
$$(a_n10^n+.....+a_2*10^2)^2+2(a_n10^n+....+a_2*10^2*78)+6084\equiv 1 \pmod4$$
so $$(a_n10^n+.....+a_2*10^2)^2+2(a_n10^n+....+a_2*10^2*78)+6083 \equiv 0 \pmod{4}$$
The first two algebraic parts of the expression must be even and 6083 is odd which implies that the sum is odd which can't be congruent to $0 \pmod4$