It will be 2017 or not

We know that $2017\equiv 1 \pmod 7$ . So, we can conclude that $4x^{3}+7y^{3}\equiv 1 \pmod 7$ $4x^{3}\equiv 1 \pmod 7$ $x^{3}\equiv 2 \pmod 7$ For $x\equiv 0,1,2,3,4,5,6 \pmod 7$ . There will be $x^{3}\equiv 0,1,6 \pmod 7$ . So, there are no solution in postive integers for $x$ . And we can conclude that there are no pairs of positive integers $(x,y)$

Since RHS is 3 modulo 4 so must be LHS.So substituting y=4k-1 we transform the equation into x^3+112k^3+21k-84k^2=506.Now since RHS is 2 modulo 7 so must be LHS.But in lhs except x^3 every term is divisible by 7.So x^3 is 2 modulo 7 which is not possible since cubic residues of 7 are 0,1,and 6.

Well I don't know if its correct or not, but I did it this way:- $4x^{3} + 7y^{3}=2017$ $(4^{1/3}x)^3 + (7^{1/3}y)^{3}=2017$

Now as 2017 is prime, the $x+y$ term in $x^{3}+y^{3}$ must be equal to $1$ . But $(4^{1/3}x+7^{1/3}y)$ can never be equal to $1$ for non negative values of x and y. Thus there are $\boxed{0}$ solutions