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Numbers with a quadratic property

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Opalg

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Feb 7, 2012
2,681
A recent Problem of the Week asked about properties of a pair of positive integers $x$, $y$ such that $2x^2+x = 3y^2+y$. But it is not obvious that any such pairs exist. So the challenge is, are there any such pairs of positive integers? If so, what is the smallest such pair? After that, what is the next smallest pair?
 
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Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
A recent Problem of the Week asked about properties of a pair of positive integers $x$, $y$ such that $2x^2+x = 3y^2+y$. But it is not obvious that any such pairs exist. So the challenge is, are there any such pairs of positive integers? If so, what is the smallest such pair? After that, what is the next smallest pair?
Hint:
As shown in the POTW thread, $x-y$ is a perfect square, say $x-y = u^2$. Then $y = x-u^2$. Use that to find and solve an equation for $x$ is terms of $u$. What condition must $u$ satisfy to ensure that $x$ is an integer?
 

castor28

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MHB Math Scholar
Oct 18, 2017
245

As a matter of fact, I was interested in that very question. The question is about finding points with integer coordinates on a hyperbola, and this is a classical problem on representation by quadratic forms. I wrote something about it here. Sorry, it's in French, but ‶the equations speak for themselves″ :)
 
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Opalg

MHB Oldtimer
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Feb 7, 2012
2,681
Congratulations to castor28 for his solution. Mine is quite similar:

Substituting $y=x-u^2$, the equation $2x^2+x = 3y^2+y$ becomes $$2x^2+x = 3(x-u^2)^2 + x - u^2 = 3x^2 - 6u^2x + 3u^4 + x - u^2,$$ $$x^2 - 6u^2x + u^2(3u^2-1) = 0,$$ $$x = 3u^2 \pm\sqrt{u^2(6u^2 + 1)}.$$ We want $x$ to be positive, so take the positive square root to get $x = 3u^2 + uv$, where $v = \sqrt{6u^2+1}$. We also want $v$ to be an integer, so we want integer solutions to the equation $v^2 = 6u^2+1$. That is a Pell-type equation. To see how to solve it, divide by $u^2$ to get $\left(\frac vu\right)^2 = 6 + \frac1{u^2}$. If $u$ is large, then $\frac1{u^2}$ is very small and so $\frac vu$ will be close to $\sqrt6$. The best rational approximations to $\sqrt6$ come from its continued fraction convergents. The helpful continued fraction calculator here gives this table:

Screenshot 2020-04-13 at 12.38.51.png

The convergents $\frac vu$ are alternately slightly larger and slightly smaller than $\sqrt6$, corresponding to solutions of $v^2 = 6u^2+1$ and $v^2 = 6u^2-1$. So from alternate rows of that table we get

$$\begin{array}{r|r|r|r}v&u&x=u(3u+v)&y=u(2u+v) \\ \hline 5&2&22&18 \\ 49&20&2180&1780 \\ 485&198&213642&174438 \\ 4801&1960&20934760&17093160\end{array}$$

After that, the numbers increase rapidly, giving an infinite sequence of positive integer solutions of $2x^2+x = 3y^2+y$. (I'm pleased to see that my numbers tally exactly with castor28 's!)