# Problem Of The Week #411 Apr 5th, 2020

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#### anemone

##### MHB POTW Director
Staff member
Here is this week's POTW:

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Suppose that the positive integers $x, y$ satisfy $2x^2+x=3y^2+y$. Show that $x-y, 2x+2y+1, 3x+3y+1$ are all perfect squares.

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#### anemone

##### MHB POTW Director
Staff member
Congratulations to the following members for their correct solution!

If $2x^2+x = 3y^2+y$ then $2(x^2-y^2) + (x-y) = y^2$ and so $(2x+2y+1)(x-y) = y^2$.
Now suppose that $p$ is a prime divisor of $x-y$. Then $p$ divides $y^2$ and hence $p$ divides $y$. Thus $p$ divides $2x+2y$, so it does not divide $2x+2y+1$. But $p$ occurs to an even power in the prime factorisation of $y^2$. Since it does not occur as a factor of $2x+2y+1$, it must occur to an even power in the prime factorisation of $x-y$. As this is true for every prime factor of $x-y$, it follows that $x-y$ is a square, say $x-y = z^2$. But then $z^2$ is a factor of $y^2$, and $2x+2y+1 = \frac{y^2}{z^2} = \left(\frac yz\right)^2$, so that $2x+2y+1$ is a square.
If we write the equation $2x^2+x = 3y^2+y$ as $3(x^2-y^2) + (x-y) = x^2$, then $(3x+3y+1)(x-y) = x^2$, and the same argument as before shows that $3x+3y+1$ is a square.