I found this question in a book of maths MCQs under the chapter "Numbers". No solution was provided within that book. I can't solve it no matter how hard I try. Please help me.
P.S. : This is my first question on stack exchange so if I made any mistake, please don't be too harsh.
The actual problem in the book is probably the following:
Prove that if $ a $ is an odd integer such that $ 11 + 11 \sqrt{11a^2 + 1} $ is rational, then $ 11 + 11 \sqrt{11a^2 + 1} $ is a perfect square.
This is indeed true. An integer which is a rational perfect square is an integer perfect square, so let $ k > 0 $ be such that $ k^2 = 11a^2 + 1 $. Our goal is to prove that $ 11(k+1) $ is a perfect square. To do this, we will show that the power of $ p $ in the prime factorization of $ k+1 $ for any prime $ p \neq 11 $ is even, and the power of $ 11 $ is odd.
We have $ 11a^2 = (k-1)(k+1) $, and since $ a $ is odd, the same is true for $ k-1, k+1 $. It follows that $ \gcd(k-1, k+1) = 1 $. If $ p \neq 11 $ is a prime dividing $ k+1 $, then the power of $ p $ in $ k+1 $ is its power in $ 11a^2 $; in other words, it is even. It follows that $ 11(k+1) $ is almost a perfect square - we are done if we show that $ k \equiv -1 \pmod{11} $. Standard theory of Pell equations yields that $ 10 + 3\sqrt{11} $ is a fundamental unit for $ \mathbf Z[\sqrt{11}] $, and we have that, defining
$$ (10 + 3 \sqrt{11})^n = a_n + b_n \sqrt{11} $$
we see that $ a_n $ is $ 2 $-periodic modulo $ 11 $, and $ b_n $ is $ 2 $-periodic modulo $ 2 $. It follows that if $ b_n $ is odd, then $ a_n $ is $ -1 $ modulo $ 11 $. This implies that $ k $ is $ -1 $ modulo $ 11 $ when $ a $ is odd, and since $ a $ is odd, we are done.