MHB Can Integer Equations Reveal Perfect Square Patterns?

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The discussion presents a problem involving positive integers x and y, requiring proof that the expressions x-y, 2x+2y+1, and 3x+3y+1 are all perfect squares given the equation 2x^2+x=3y^2+y. Several members successfully solved the problem, including kaliprasad, castor28, and Opalg. The thread emphasizes the importance of following the Problem of the Week guidelines for submissions. The solutions provided demonstrate various approaches to proving the perfect square conditions. Overall, the thread showcases mathematical exploration of integer equations and their properties.
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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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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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Congratulations to the following members for their correct solution!(Cool)

1. kaliprasad
2. castor28
3. Opalg

Solution from Opalg:
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.
 
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