QUESTION

# Find the general solution in positive integers of ${x^2} - 5{y^2} = 1$.

Hint: Assume, x and y as $x = 2n,y = 2n + 1$, since the R.H.S is odd, so both x and y cannot be even or odd. put in the equation given and solve.

We have been given the equation- ${x^2} - 5{y^2} = 1$
Now, since the R.H.S is odd, so assume either x to be an even integer and y as an odd integer, or vice- versa.
So, assuming $x = 2n,y = 2n + 1$ and put these in the given equation, ${x^2} - 5{y^2} = 1$, we get-
${(2n)^2} - 5{(2n + 1)^2} = 1$

Solving further we get,
$\Rightarrow {(2n)^2} - 5{(2n + 1)^2} = 1 \\ \Rightarrow 4{n^2} - 5(4{n^2} + 1 + 4n) = 1 \\ \Rightarrow 4{n^2} - 20{n^2} - 5 - 20n = 1 \\ \Rightarrow - 16{n^2} - 20n - 6 = 0 \\ \Rightarrow 16{n^2} + 20n + 6 = 0 \\$

So, we get a quadratic equation in n.
Now, finding the roots of the equation we get-
$16{n^2} + 20n + 6 = 0 \\ \Rightarrow 16{n^2} + 8n + 12n + 6 = 0 \\ \Rightarrow 8n(2n + 1) + 6(2n + 1) = 0 \\ \Rightarrow (8n + 6)(2n + 1) = 0 \\ \Rightarrow n = \dfrac{{ - 3}}{4},\dfrac{{ - 1}}{2} \\$

So, now put the values of n in $x = 2n,y = 2n + 1$.
For $n = - \dfrac{3}{4}$, we get $x = 2 \times \dfrac{{ - 3}}{4} = \dfrac{{ - 3}}{2},y = 2n + 1 = \dfrac{{ - 3}}{2} + 1 = - \dfrac{1}{2}$
For $n = - \dfrac{1}{2}$, we get $x = 2 \times \dfrac{{ - 1}}{2} = - 1,y = 2n + 1 = - 1 + 1 = 0$
We can see, x = -3/2 and y = -1/2, satisfies the equation, ${x^2} - 5{y^2} = 1$.
And, x =-1 and y = 0, also satisfies the equation, ${x^2} - 5{y^2} = 1$.
Hence, the general solution in positive integer is-
X = -3/2, y = -1/2 and x = -1, y = 0.

Note – Whenever such types of questions appear, always see whether the RHS is odd or even, accordingly make the assumptions for x and y. Here, we have assumed, $x = 2n,y = 2n + 1$, and then solved for the general solution in positive integers.