Given the equation

$\large \color{#D61F06}{4x^{3}}+\color{#3D99F6}{7y^{3}}=\color{#69047E}{2017}$

How many pairs of positive integers $(x,y)$ that satisfy the equation?

3
2
0
5
1
4

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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 $(x,y)$