Solve in positive integers:
$4 x^{2} - y^{2} = 315$ .

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Solution

We know that $a^{2} - b^{2} = (a + b) (a - b)$
$=> (2 x + y) (2 x - y) = 315$
$315 = 1 \times 315$
$= 3 \times 105$
$= 5 \times 63$
$= 7 \times 45$
$= 9 \times 35$
$= 15 \times 21$
We know $2 x + y > 2 x - y$ Since $x, y > 0$
By comparision
1. $2 x + y = 315, 2 x - y = 1$ $=> x = 79; y = 157$
2. $2 x + y = 105, 2 x - y = 3$ $=> x = 27; y = 51$
3. $2 x + y = 63, 2 x - y = 5$ $=> x = 17; y = 29$
4. $2 x + y = 45, 2 x - y = 7$ $=> x = 13; y = 19$
5. $2 x + y = 35, 2 x - y = 9$ $=> x = 11; y = 13$
6. $2 x + y = 21, 2 x - y = 15$ $=> x = 9; y = 3$

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