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How many integer solutions for (a, b) satisfied the following equation

a^2+b^2=ab^2

$a^2+b^2=ab^2$

$(a, \mbox { } b) = (2, \mbox {} \pm \mbox { } 2)$

$​​​​(a, \mbox { } b) = ( \pm \mbox { } 2, \mbox { } 2)$

        $(a, \mbox { } b) = (0, \mbox { } 0)$

Hope this was helpful . . . ^^

 !

Is the expression on the right supposed to be (ab)²  or  a • b²   ?  

Fist thing I did was graph it.

https://www.geogebra.org/classic/svfrstbf

a^2+b^2=ab^2

its symetrical so I want to look at the top half   b>0 (then I will double the number of answers)

$a^2=ab^2-b^2\\ a^2=b^2(a-1)\\ b=\frac{a}{\sqrt{a-1}}$

a has to be greater than 1,  

For b to be an integer  

a^2 = k (a-1)     where k is an integer

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yea I don't know.

There are at least 2 answers.