## Square or Cube

Find all positive integral solutions of $x$ and $y$ to $x^2y-y^3=105$.
Remember, x and y are positive integers.
Source: mathcontest.olemiss.edu 9/21/2009

SOLUTION
$x^2y-y^3=105$
$y\left (x^2-y^2\right )=105$
$y\left (x+y\right )\left (x-y\right )=105$
$y\left (x+y\right )\left (x-y\right )=3\times 5\times 7$
Since $x$ and $y$ are positive, $\left (x-y\right )$ must be positive in order to make the above product positive. This implies that $x>y$.

Also, since 105 is odd, all three factors $y, \left (x+y\right ), \textup{ and }\left (x-y\right )$ must be odd, which implies that $x$ is even.

So let’s try the three possible values for $y$: 3, 5, or 7.

#### $\mathbf{y=3}$

 y x x + y x – y y(x + y)(x – y) 3 4 7 1 3(7)(1)=21 6 9 3 3(9)(3)=81 8 11 5 3(11)(5)=165 >105

Thus, there is no solution when $y=3$.

$\mathbf{y=5}$

 y x x + y x – y y(x + y)(x – y) 5 6 11 1 5(11)(1)=55 8 13 3 5(13)(3)=195 >105

Thus, there is no solution when $y=5$.

$\mathbf{y=7}$

 y x x + y x – y y(x + y)(x – y) 7 8 15 1 7(15)(1)=105

There is a solution when $y=7$ and $x=8$.

Check
$x^2y-y^3=105$
$8^2\times 7-7^3=105$?
$64\times 7-343=105$
$448-343=105$
$105=105$. YES.

Answer: $x=8,\;y=7$.