State True or False. The number of four digit numbers which are perfect squares and in which the first two digits are equal to each other and the last two digits are equal to each other is 2

Such a number is $$\overline{aabb} = 1100a + 11b = 11(100a + b)$$, so 11 divides it and $$100a + b$$ must be 11 times a square, i.e. $$a + b = 11$$ and $$\overline{aabb} = 121(9a + 1)$$. The only digit $$a$$ making $$9a + 1$$ a perfect square is $$a = 7$$, giving $$7744 = 88^2$$. There is only 1 such number, so the statement is False.