Question

The number of ordered pairs of positive integers x, y such that $x^2+3y$ and $y^2+3x$ are both perfect squares is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

Let $y\ge x$. Then the following inquality must also hold :

$(y+2{)}^2>y^2+3x>y^2$

Therefore we must have that $y^2+3x=(y+1{)}^2$ or:

$y^2+3x=y^2+1+2y$

$3x=2y+1$

$y=(3x-1)/2$ ......(i)

It is also given that

$x^2+3y=x^2+9/2x-3/2$

is a perfect square. Again, we can deduce another inequality from this statement :

$(x+3{)}^2>x^2+9/2x-3/2>x^2$

now, the only equations we have to solve are :

$x^2+9x/2-3/2=(x+1{)}^2$ and

$x^2+9x/2-3/2=(x+1{)}^2$

or

$9x/2-3/2=2x+1$ ..... (ii)

$9x/2-3/2=4x+4$ .......(iii)

which give us $x=1$ and $x=11$ respectively and by putting these values of x in eq. (i) we get, $y=1$ and $y=16$

now, again by assuming $x\ge y$, we get $x=16$, $y=11$.

So, the required ordered pairs are $(1,1)(11,16)(16,11)$