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I found the following equation form an answer written for a question.

$$2013 y^2 -xy -4026 x=0$$

But I'm confused that can I really learn how to find the positive integer solutions for $x,y$ with having only the middle school mathematics knowledge. How do I find the positive integer solutions for this equation?

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Tharindu Sathischandra
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2 Answers2

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Your equation is equivalent to the following $$x=\frac{2013y^2}{y+4026}$$ so you can restrict yourself on looking for $y$ such that this fraction is an integer (note that, for positive $y$, automatically $x$ is positive). Call $y+4026=t$ (so that $y=t-4026$). Recall that, if $y \ge 1$, then $t \ge 4027$. Then $$x=\frac{2013(t-4026)^2}{t} = 2013 t - 2^2 \cdot 2013^2+ \frac{2^2 \cdot 2013^3}{t}$$ The RHS is an integer if and only if $t$ divides $2^2 \cdot 2013^3$. Factor that number: $$2^2 \cdot 2013^3=2^2 \cdot 3^3 \cdot 11^3 \cdot 61^3$$ So, now you have to find all divisors $t$ of this number satisfying $t \ge 4027$ (there are a lot of them). For every such value of $t$, you can solve $y= t-4026$, and then $x = \frac{2013y^2}{t}$.

Crostul
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    For a table of all divisors of $2^2 \cdot 2013^3$, see this link. – Crostul Mar 04 '16 at 11:12
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    Using a computer, I found that the smallest solution is $(x,y)=(50325,330)$, hence I hardly doubt that this can be found using only pen and paper without making mistakes in computations. Moreover, there exist 139 solutions (a lot of them). – Crostul Mar 04 '16 at 11:25
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    If I just wanted to find some solution by hand, I would choose $t=3 \cdot 2013, y=2013, x=2013^3/3$. Cubing $2013$ is not so hard by expanding $(2000+13)^3$ – Ross Millikan Mar 04 '16 at 20:04
  • @Crostul: Finding this solution seems trivial with your approach. Determining that it's minimal seems only a little tedious. Finding them all is, I agree, unreasonable for pencil and paper. – Brian Tung Mar 04 '16 at 20:04
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An scholar way can be the following:

Put $a=2013=3\cdot 11\cdot 61$; we see there are $(1+1)(1+1)(1+1)=8$ possible factors.

$ay^2-xy-2ax=0\Rightarrow \color{red}{a|xy}\qquad (1)$.

$ay^2-xy-2ax=0\Rightarrow 2ay=x\pm \sqrt{x^2+8a^2x}\Rightarrow \color{red}{x^2+8a^2x=z^2}\Rightarrow x^2+8a^2x=w^2x^2\Rightarrow \color{red}{x|8a^2}\qquad (2)$.

So $0<x|2^3\cdot3^2\cdot11^2\cdot61^2$ hence there are at most $(3+1)(2+1)(2+1)(2+1)=108$ possibilities for solution $x$.

(There are not infinitely many solutions!)

From $(1)$ and $(2)$ one has $xy=2013k$ and $x^2+8a^2x=\square$ (= square) We try $k\in \{1,2\}$

►$k=1$

The possible solutions are $(x,y)\in \{(1,2013),(3,671),(11,183),(33,61)\}$ and the opposite $(y,x)$ with the four given possibilities. Here we examine, using $(2)$, how squares, with eventual solutions can agree $$\begin{cases} 1+8(2013^2=\square\\9+24(2013^2=\square\\11^2+8\cdot 11(2013)^2=\square\\33^2+8\cdot 33(2013)^2=\square\\(2013)^2(1+8\cdot2013)=\square\\(671)^2(1+8\cdot 671\cdot 3^2)=\square\\(183)^2(1+8\cdot 183\cdot11^2)= \square\\(61)^2(1+8\cdot61\cdot 33^2=\square\end{cases}$$

(the reason of the factorization in four last possibilities is to simplify the calculation).

None of these eight possibilities is square; then they are discarded.

►$k=2$

The $16$ possibilities of eventual solution for $xy=4026$, like for $k=1$, give not squares so they must be discarded.

The smallest solution, in case it exists, seems to be large. Without computer I doubt a student with the middle school math knowledge could find out a solution.

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