A number theory problem by Ilham Saiful Fauzi

x^3 +4y^3(x^3 -2) =827 Now consider the equation mod 4, You’ll get x^3=3 (mod 4). Now try 1^3=1; 2^3=0; 3^3=27=3 mod 4. Therefore x=3. Plugging back x=3 into the original equation we get 4y^3(27-2)=827-27=800 or y^3=800/4*25 or y^3=8 hence y=2.

From the equation, x is clearly odd. Define x = 2r +1 and expand, so that x^3 = 8r^3 +12r^2 + 6r +1. Substituting into the equation, transpose 1 to the left, and divide by 2. Defining k = 4r^3 + 6r^2 + 3r, the equation becomes: k + (2y^3)*(2k -1) = 413. since y=1 is impossible, we try y = 2. The previous equation becomes 33k = 429, and k = 13. From the definition of k, we see that r =1, and x = 2r +1 = 3, so x=3, y = 2, and x + y = 5. Ed Gray

x^3(1+4y^3)-8y^3=825+2 x^3(1+4y^3)-8y^3-2=825 x^3(1+4y^3)-2(4y^3+1)=825 (1+4y^3)(x^3-2)=825 825=5^2×3×11 (1+4y^3)(x^3-2)=(3×11)(5×5) x=3 y=2

We can write the given equation as follows $x^{3}+4x^{3}y^{3}-8y^{3}=827$ $\Rightarrow x^{3}+4x^{3}y^{3}=8y^{3}+827$ $\Rightarrow x^{3}(1+4y^{3})=8y^{3}+827$ $\Rightarrow x^{3}=\frac{8y^{3}+827}{4y^{3}+1}$ $\Rightarrow x^{3}=2+\frac {825}{1+4y^{3}}$ Since $x^{3}$ is an integer then $1+4y^3$ must be factor of $825$ which are $1,3,5,11,15,25,33,55,75,165,275$ or $825$ . The only possibility is $x^{3}=2+25\Rightarrow x=3$ and, $1+4y^{3}=\frac {825}{25}=33\Rightarrow y=2$ Thus $x+y=3+2=\boxed{5}$