We write
\[ (x+2y)^2 \; = \; x^2 + 4y^2 + 4xy \; = \; (2xy-7)^2 + 4xy \; = \; 4x^2y^2 - 24xy + 49 \; = \; (2xy-6)^2 + 13 \]
If we write \(a=x+2y\) and \(b=xy-3\) we have
\[ 13 \; = \; a^2 - 4b^2 \; = \; (a + 2b)(a - 2b) \]
We have the following possible factorisations:
\[ \begin{array}{|c|c|c|c|} \hline
a+2b & a-2b & a=x+2y & b=xy-3 \\ \hline
13 & 1 & 7 & 3 \\
1 & 13 & 7 & -3 \\
-13 & -1 & -7 & -3 \\
-1 & -13 & -7 & 3 \\ \hline
\end{array} \]
There are four cases to consider:

If \(x+2y=7\) and \(xy-3=3\) then \(6 = xy = (7-2y)y\), so that
\[ 0 \; = \; 2y^2 - 7y + 6 \; = \; (2y -3)(y-2) \]
and hence the only integer solution is \(y=2\), with corresponding \(x=3\).

If \(x+2y=7\) and \(xy-3=-3\) then \(0 = xy = (7-2y)y\), so the only integer solution is \(y=0\), with corresponding \(x=7\).

If \(x+2y=-7\) and \(xy-3=-3\) then \(0 = xy = -(7+2y)y\), so the only integer solution is \(y=0\), with corresponding \(x=-7\).

If \(x+2y=-7\) and \(xy-3=3\) then \(6 = xy = -(7+2y)y\), so that
\[ 0 \; = \; 2y^2 + 7y + 6 \; = \; (2y + 3)(y + 2) \]
and hence the only integer solution is \(y=-2\), with corresponding \(x=-3\).

Thus there are four solutions: \((3,2)\), \((-3,-2)\), \((7,0)\) and \((-7,0)\).

My answer is (for all integral solutions only).
Ordered pairs (x, y) = (7, 0), (-7, 0), (3, 2), (-3, -2)

Why is this so?
We can decompose x^2 + 4y^2 = (x^2 + 4xy + 4y^2) - 4xy = (x + 2y + 2(sqrt(xy))) (x + 2y - 2 (sqrt(xy))) = 4(x^2)(y^2) - 24xy + 49

It is trivial putting xy = 0 which implies that either x or y is zero. If x is zero, then y = 7/2 or -7/2 which is not an interger. If y is zero, then x = 7 or -7.

Use the discriminant and we determine that the right-hand side of the polynomial is factorable but by complex imaginary numbers which is 4(xy)^2 - 24xy + 49 = (2xy - (6 + i sqrt(13)) (2xy - (6 - i sqrt(13)) = (2xy - 6)^2 + 13.

Further info: Regarding the factorized form of the right-hand side of the equation (on the last part), if you're going to observe the behavior especially when operations are applied, this a square number added always to 13. But since the difference of the perfect squares are increased by 2 which implies that there is only two pair of perfect squares of consecutive integers (the other is negative) that satisfy the condition. So the only pair that satisfy these conditions is 36 and 49.

We equate 49 (since it is greater). This implies that xy = 6 or 0. We already discussed for xy = 0 but for xy = 6 can be decomposed for 6 = (-1)(-6) = (-2)(-3) = (1)(6) = (2)(3). And by checking these values, hence, the answer.

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TopNewestWe write \[ (x+2y)^2 \; = \; x^2 + 4y^2 + 4xy \; = \; (2xy-7)^2 + 4xy \; = \; 4x^2y^2 - 24xy + 49 \; = \; (2xy-6)^2 + 13 \] If we write \(a=x+2y\) and \(b=xy-3\) we have \[ 13 \; = \; a^2 - 4b^2 \; = \; (a + 2b)(a - 2b) \] We have the following possible factorisations: \[ \begin{array}{|c|c|c|c|} \hline a+2b & a-2b & a=x+2y & b=xy-3 \\ \hline 13 & 1 & 7 & 3 \\ 1 & 13 & 7 & -3 \\ -13 & -1 & -7 & -3 \\ -1 & -13 & -7 & 3 \\ \hline \end{array} \] There are four cases to consider:

If \(x+2y=7\) and \(xy-3=3\) then \(6 = xy = (7-2y)y\), so that \[ 0 \; = \; 2y^2 - 7y + 6 \; = \; (2y -3)(y-2) \] and hence the only integer solution is \(y=2\), with corresponding \(x=3\).

If \(x+2y=7\) and \(xy-3=-3\) then \(0 = xy = (7-2y)y\), so the only integer solution is \(y=0\), with corresponding \(x=7\).

If \(x+2y=-7\) and \(xy-3=-3\) then \(0 = xy = -(7+2y)y\), so the only integer solution is \(y=0\), with corresponding \(x=-7\).

If \(x+2y=-7\) and \(xy-3=3\) then \(6 = xy = -(7+2y)y\), so that \[ 0 \; = \; 2y^2 + 7y + 6 \; = \; (2y + 3)(y + 2) \] and hence the only integer solution is \(y=-2\), with corresponding \(x=-3\).

Thus there are four solutions: \((3,2)\), \((-3,-2)\), \((7,0)\) and \((-7,0)\).

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My answer is (for all integral solutions only). Ordered pairs (x, y) = (7, 0), (-7, 0), (3, 2), (-3, -2)

Why is this so? We can decompose x^2 + 4y^2 = (x^2 + 4xy + 4y^2) - 4xy = (x + 2y + 2(sqrt(xy))) (x + 2y - 2 (sqrt(xy))) = 4(x^2)(y^2) - 24xy + 49

It is trivial putting xy = 0 which implies that either x or y is zero. If x is zero, then y = 7/2 or -7/2 which is not an interger. If y is zero, then x = 7 or -7.

Use the discriminant and we determine that the right-hand side of the polynomial is factorable but by complex imaginary numbers which is 4(xy)^2 - 24xy + 49 = (2xy - (6 + i sqrt(13)) (2xy - (6 - i sqrt(13)) = (2xy - 6)^2 + 13.

Further info: Regarding the factorized form of the right-hand side of the equation (on the last part), if you're going to observe the behavior especially when operations are applied, this a square number added always to 13. But since the difference of the perfect squares are increased by 2 which implies that there is only two pair of perfect squares of consecutive integers (the other is negative) that satisfy the condition. So the only pair that satisfy these conditions is 36 and 49.

We equate 49 (since it is greater). This implies that xy = 6 or 0. We already discussed for xy = 0 but for xy = 6 can be decomposed for 6 = (-1)(-6) = (-2)(-3) = (1)(6) = (2)(3). And by checking these values, hence, the answer.

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Sorry, but \(xy\) does not have to be a square. \(x=3\) and \(y=2\) is a solution.

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Yes.. I edited... I just noticed before I edit this.. Thanks... The factoring mislead me..

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