Consider the following equation: $$|x+y^2|+|x-y^2|+|y+x^2|+|y-x^2|=a$$ I'm looking for the method for solving some problems regarding this equation, namely: 1) prove that if $a=2015$, then the equations has no solutions such that $x,y\in\mathbb{Z}$; 2) prove that if $a=2112$, then the equations has even number of solutions such that $x,y\in\mathbb{Z}$. I currently have no idea how to deal with such an equation.
1 Answer
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To prove the first part (i.e. that there is no solution for $a=2015$), note that the left hand side is always an even number. The reason is that the left hand side has 4 terms and the parity of all the terms is the same (they are either all even or all odd). The sum of 4 odd numbers is even, so is the sum of 4 even numbers.
To see that the number of solutions for $a=2112$ (or any other $a \neq 0$) is even, note that for any solution $(x, y)$, you can flip the sign of $x$ and get a new valid solution; and the same for $y$.
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$\begingroup$ Amazing. So it was THAT simple? Thank you so much! $\endgroup$– user263286Dec 6, 2015 at 0:48