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Find the number of integer solutions to the equation y^2 − 6y + 2x^2 + 8x = 367

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First, let's rearrange the equation by completing the square for y and x.

(y-3)^2 -9 + 2(x+2)^2 -8 = 367

Next, simplify and isolate the squared terms.

2(x+2)^2 + (y-3)^2 = 384

Divide both sides by 2 to obtain

(x+2)^2 + (y-3)^2 / 2= 192

To find integer solutions, we can try different values of x and solve for y. If y turns out to be an integer, then we have found a solution. It might also be helpful to note that the sum of two squares is an integer only if both squares are integer (or zero).

Let's try x = 0, 1, 2.

For x = 0, we have

(y-3)^2 = 384

This has no integer solution.

For x = 1, we have

(y-3)^2 = 382

This also has no integer solution.

For x = 2, we have

(y-3)^2 = 378

This gives us two integer solutions:

y-3 = ±6√3

y = 3 ± 6√3

So, the equation has two integer solutions:

(2, 3 + 6√3) and (2, 3 − 6√3)
User Igor Tkachenko
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