Count the ordered pairs again-2

Number Theory Level 5

Find number of ordered pairs $(x,y)$ of non negative integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares.

$\bullet$ Enter $777$ as answer if there are infinte number of ordered pairs.

$\bullet$ Ordered pair means $(11,12$ and $(12,11)$ are considered different.

This is a part of Number Theory Plus

The answer is 777.

2 solutions

Ravi Dwivedi
Jul 17, 2015

The equalities $x^2+3y \geq (x+2)^2, y^2+3x \geq (y+2)^2$ cannot hold simultaneously because summing them yields $0\geq x+y+8$ which is false because $x$ and $y$ are non negative integers.

Hence atleast one of $x^2+3y < (x+2)^2, y^2+3x < (y+2)^2$ is true.

Without loss of generality assume that $(x^2+3y < (x+2)^2 \implies x^2+3y=(x+1)^2$ hence $3y=2x+1$

Then $x=3k+1$ and $y=2k+1$ for some integer $k \geq 0$

So $y^2+3x=4k^2+13k+4$

If $k>5$ then $(2k+3)^2<4k^2+13k+4<(2k+4)^2$

So $y^2+3x$ cannot be a square. It is easy to check that for $k\in \{1,2,3,4\}, y^2+3x$ is not a sqaure but for $k=0,y^2+3x=4=2^2$ .

Therefore the only solution is $(x,y)=(1,1)$

Moderator note:

Your first error comes about when you say

$x^2 + 3y < (x+2)^2 \Rightarrow x^2 + 3y = (x+1) ^2$

You forgot that you allowed for $y$ to be non-negative.

The other error, as pointed out by Isaac, is that $k=5$ yields $(x,y) = (11,16)$ .

What about $(11,16)$ and $(16,11)$ ?

This question has been done before here by Calvin.

Also since we're dealing with non negative integers. Let $x=0$ and $y=3^{2k+1}$

Since this works for all natural values of $k$ the answer should be $\boxed{777}$

Isaac Buckley - 5 years, 11 months ago

Thanks. I have updated the answer to 777 since $(0, 3 k^2 )$ is a solution.

Calvin Lin Staff - 5 years, 11 months ago

But I could not point out mistake in my solution. Please do that

Ravi Dwivedi - 5 years, 11 months ago

The case where $k=5$ seems to be missing. Also the answer should be changed to $\boxed{777}$ .