I decided to have a go and here is how I went. Let me know if I made any mistakes
Consider the Diophantine equation
[tex]x^3+117y^3=5[/tex]
Choose mod 5.
The equation tells us that [itex]5|x^3 \Rightarrow 5^3|x^3 \Rightarrow x = 5X[/itex]. Therefore
[tex]125\cdot X^3 +117y^3 = 5[/tex]
By the same procedure as above, this equation tells us that [itex]5|y^3 \Rightarrow y = 5Y[/itex]. Therefore
[tex]125\cdot X^3 + 117\cdot 5Y^3 = 5[/tex]
Divide through by 5 and we have
[tex]25\cdot X^3 + 117\cdot Y^3 = 1[/tex]
This equation now tells me that
[tex]25X^3 \equiv 1(\mod 5)[/tex]
But [itex]25\equiv 0 (\mod 5)[/itex]. Hence there is no such [itex]0,1,2,3,4[/itex] such that [itex]X^3\equiv 1(\mod 5)[/itex].
However, if we had checked Y first we would have found
[tex]117Y^3 \equiv 1(\mod 5)[/tex]
which implies that
[tex]Y^3 \equiv 2(\mod 5)[/tex]
since [itex]117\equiv 2 (\mod 5)[/itex]. And since if we let [itex]Y = 3[/itex] then
[tex]Y^3 = 3^3 = 27 \equiv 2(\mod 5)[/tex]
then this tells us that there is a solution. But since it failed for X there is no solution.