Prove that the Pell's Equation $x^2 −Dy^2 = 1$ always has a solution where $y$ is a multiple of $41$ $D$ is a positive integer that is not a perfect square
Recently I am taking a introductory number theory course and I met this question right after we learned Pell's equation and Diophantine Approximation. However, I can't see a connection between those 2 topics and this question. 
I was trying to assume that $ y = 41k$ where k is an positive integer and substitute it into the equation and I hoped eventually this will simplify to an equation that conforms the form of the Pell's equation which is $x^2-Dy^2=1$. However I did not get any from there.
Also I tried to approach this problem from the Pell's Equation Theorem. Then I found it is impossible to get anything useful from expanding $(x+y{\sqrt D})^k$ plus I cannot determine the smallest solution for it because I don't know the value of D.
Could someone help me on this question? Thank you!

There is an old parametric solution for Pell equation that says, if x, y and D are certain functions of a parameter such as $m$ , there can be infinite solutions"
We rewrite equation as:
$x^2-1=Dy^2$
$1$ is odd and number of terms on LHS is even so one of terms must be odd. Suppose $x^2$ is odd and we have:
$x=2m^2+1$
$(2m^2+1)^2-1=D y^2$
$4m^2(m^2+1)=D.y^2$
So we must have:
$y^2=4m^2$ ⇒ $y=2m$
${D=m^2+1}$
So m can have any value in $\mathbb Z$, incuding all multiple of $41$.
For the smallest solution you can let $m=1$, then we have:
$D=1^2+1=2$
$x=2\times 1^2 +1=3$
$y=2\times 1=2$
If you want y a multiple of $41$, just let $m=41$, then:
$(x, y, D)= (3363, 82, 1682)$
