\(x^2-2y^2=1\) and \(x+2y=1\). From the second equation, we get \(x=-2y+1\). Substituting that into the first equation gives \( (-2y+1)^2-2y^2=1\Rightarrow 4y^2-4y+1-2y^2=1\Rightarrow 2y^2-4y+1=1\). Subtracting \(1\) from both sides gives \(2y^2-4y=0\), and dividing both sides by \(2\) gives \(y^2-2y=0\). Factoring the left side shows that \(y(y-2)=0\), so \(y=\{2,0\}\) and \(x=\{-3,1\}\). The two solution pairs that we get are \( (-3,2) \space\text{and}\space (1,0)\).

\(x^2-2y^2=-1\) and \(x+2y=-1\). From the second equation, we get \(x=-2y-1\). Substituting that into the first equation gives \( (-2y-1)^2-2y^2=-1\Rightarrow 4y^2+4y+1-2y^2=-1\Rightarrow 2y^2+4y+1=-1\). Adding \(1\) to both sides gives \(2y^2+4y+2=0\), and dividing both sides by \(2\) gives \(y^2+2y+1=0\). The left side is actually \( (y+1)^2\), so \( (y+1)^2=0\). The only solution here is \(y=-1\) and \(x=1\), and the pair that we get is \( (1,-1)\).

We do not need to consider neither if \(x+2y=\frac{a}{b}\) (where \(a\) and \(b\) are integers and \(|a|\neq|b|)\) nor if \(x^2-2y^2=\frac{a}{b}\) because both will have non-integer solution pairs (Basically, in this case, if \(x\) is an integer, \(y\) is not, and if \(y\) is an integer, \(x\) is not). Therefore, the only integer solution pairs are \(\boxed{(1,0),(1,-1)\space\text{and}\space(-3,2)}\).

i loved the approach....... so you mean if I replace that "1" over there by an integer "n" then all I had to do is split it according to fundamental theorem of arithmetic and do the rest?... :)

I am not going to post a solution since that would copy Jeff's solution, but a motivation for his solution is to note that the coefficient pattern is 1:2 and 2:4 so it could be "factored". Then we see that the ratio of the terms is also the same and we try to factor. We continue by noting we have two integers multiplying to 1, consider cases, and done.

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## Comments

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TopNewestFirst, rewrite

\(x^3+2x^2y=2xy^2+4y^3+1\)

as

\(x^3+2x^2y-2xy^2-4y^3=1\).

We find that

\(x^3+2x^2y-2xy^2-4y^3=x^2(x+2y)-2y^2(x+2y)=(x^2-2y^2)(x+2y)\),

so

\( (x^2-2y^2)(x+2y)=1\).

We check \(2\) cases:

\(x^2-2y^2=1\) and \(x+2y=1\). From the second equation, we get \(x=-2y+1\). Substituting that into the first equation gives \( (-2y+1)^2-2y^2=1\Rightarrow 4y^2-4y+1-2y^2=1\Rightarrow 2y^2-4y+1=1\). Subtracting \(1\) from both sides gives \(2y^2-4y=0\), and dividing both sides by \(2\) gives \(y^2-2y=0\). Factoring the left side shows that \(y(y-2)=0\), so \(y=\{2,0\}\) and \(x=\{-3,1\}\). The two solution pairs that we get are \( (-3,2) \space\text{and}\space (1,0)\).

\(x^2-2y^2=-1\) and \(x+2y=-1\). From the second equation, we get \(x=-2y-1\). Substituting that into the first equation gives \( (-2y-1)^2-2y^2=-1\Rightarrow 4y^2+4y+1-2y^2=-1\Rightarrow 2y^2+4y+1=-1\). Adding \(1\) to both sides gives \(2y^2+4y+2=0\), and dividing both sides by \(2\) gives \(y^2+2y+1=0\). The left side is actually \( (y+1)^2\), so \( (y+1)^2=0\). The only solution here is \(y=-1\) and \(x=1\), and the pair that we get is \( (1,-1)\).

We do not need to consider neither if \(x+2y=\frac{a}{b}\) (where \(a\) and \(b\) are integers and \(|a|\neq|b|)\) nor if \(x^2-2y^2=\frac{a}{b}\) because both will have non-integer solution pairs (Basically, in this case, if \(x\) is an integer, \(y\) is not, and if \(y\) is an integer, \(x\) is not). Therefore, the only integer solution pairs are \(\boxed{(1,0),(1,-1)\space\text{and}\space(-3,2)}\).

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i loved the approach....... so you mean if I replace that "1" over there by an integer "n" then all I had to do is split it according to fundamental theorem of arithmetic and do the rest?... :)

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I am not going to post a solution since that would copy Jeff's solution, but a motivation for his solution is to note that the coefficient pattern is 1:2 and 2:4 so it could be "factored". Then we see that the ratio of the terms is also the same and we try to factor. We continue by noting we have two integers multiplying to 1, consider cases, and done.

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thts a nice approach.......... could you please generalize that observation and let in a more general rule for factorization :)...(need help)

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@Daniel Liu , we want you as a problem writer for Proofathon. Email proofathon@gmail.com if you're interested.

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Good for you @Daniel Liu! :D

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