Find the number of pairs $(x,y) \in \mathbb{Z}$ such that

$x^3+y^4=7$

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The answer is 0.

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First note that there is no solution in both positive $(x,y)$ .So, if there exists any solution then exactly one of them should be negative.

As $y^4 >0$ so $x$ should be negative.Let's assume $x=-z$ .Then our equation becomes

$y^4 -z^3=7 \implies y^4=z^3+7$

Case(1)When $z$ is evenAssume $z=2k$ then $y$ will also become odd Assume $y=2l+1$

Substituting this in the equation $(2l+1)^4=(2k)^3+7$

Now using binomial expansion $8l'+1 =8k^3+7 \implies 8(l'-k^3)=6$ which is clearly a contradiction.So, there is no even value of $z$ exists.

Case(2)When $z$ is oddNow consider $z=2^l k+1$ where $k$ is strictly odd. So, $y$ will be even.

Substituting this we get $y^4=(2^l k+1)^3 + 7 \\ =2^{8l} k^3+3.2^{4l} k^2+3.2^{l} k+1+7 \\=2^{8l} k^3+3.2^{4l} k^2+3.2^{l} k+8 \\ =8(2^{8l-3} k^3+3.2^{4l-3} k^2+3.2^{l-3} k+1)$

As $y$ is even so, $16 | y^4$ .Here , $8 | y^4$ which is not possible because rest of the part is completely odd.

So, there doesn't exists any odd value.

Therefore, there is no integal solution of the given equation.