Question 9
The number of ordered pairs $$(x,y)$$ of integers such that $$x-y^2=4$$ and $$x^2+y^4=26$$ is
Solution
From $$x-y^2=4$$, we get $$x=4+y^2$$. Substitution into the second equation gives $$(4+y^2)^2+y^4=26$$, or $$y^4+4y^2-5=0$$. This factors as $$(y^2+5)(y^2-1)=0$$, so the integer values are $$y=1$$ and $$y=-1$$, with $$x=5$$ in both cases. Thus there are $$2$$ ordered pairs.
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