If $f(x,y)=10^5x^5+50000x^4y+10^4x^3y^2+10^3x^2y^3+50xy^4$, then is it possible that $f(x_1,y_1)=f(x_2,y_2)$ for $x_1\neq x_2$ and $y_1\neq y_2$ for all x,y belongs to Z+?
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Yes, it is possible. For example $$f(0,0)=0$$ but the polynomial can have many zeros – Maximilian Janisch Dec 24 '19 at 14:10
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@MaximilianJanisch : But note that you would have to demonstrate that there is a root with both coordinates nonzero – MPW Dec 24 '19 at 14:13
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@MPW It is easy to construct such a polynomial by choosing for example $y=1$ then you only need to choose $a,b,c,d\in\mathbb R$ such that $f(x,1)=ax^5+bx^4+cx^3+dx=0$ for some $x\neq0$ which is not hard (for example $a+b+c+d=0$ and $x=1$.) – Maximilian Janisch Dec 24 '19 at 14:16
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It seems to me that OP is asking if this is possible for all given coefficients. That’s not as clear to me, but I agree that it is probably true since there are no further constraints – MPW Dec 24 '19 at 14:18
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https://math.stackexchange.com/q/1509175/631742 – Maximilian Janisch Dec 26 '19 at 12:42
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Yes, the polynomial does not have to be one-to-one.
For example $$ f(x,y)=x^5-5x^4y $$ have solutions $$ f(0,0)=f(5,1)=0$$
Mohammad Riazi-Kermani
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