Simultaneous Equations Shakespearean Edition

Relevant wiki: Fermat's Last Theorem

\[\begin{cases}\begin{array}{} 5x^3+3y^3=3(x+y)(x+1)(y+1)-1 & \qquad(1)\\ x^3+2y^3=1 & \qquad(2)\end{array}\end{cases}\]

We perform the following manipulation: $\begin{aligned} (2)&\implies 2x^3+4y^3=2 \\ &\implies 3x^3+5y^3=x^3+y^3+2\qquad(3) \end{aligned}$

$\begin{aligned} (1)+(3)\implies 8x^3+8y^3&=x^3+y^3+1+3(x+y)(x+1)(y+1) \\ &=x^3+y^3+1+3(x+y)(xy+x+y+1) \\ &=x^3+y^3+3xy(x+y)+1+3(x+y)(x+y+1) \\ &=(x+y)^3+1+3(x+y)(x+y+1) \\ &=(x+y+1)^3 \end{aligned}$

$\therefore 8x^3+8y^3=(x+y+1)^3\implies(2x)^3+(2y)^3=(x+y+1)^3$

Since $x$ and $y$ are integers, by Fermat's Last Theorem (also see Fermat's Last Theorem equivalent statements ), this equation has no non-trivial integer solutions.

Thus, one of $2x$ , $2y$ or $x+y+1$ has to be equal to $0$ .

Case 1: If $2x=0\implies x=0$

Then $(2y)^3=(0+y+1)^3\implies 2y=y+1\implies y=1$ However, $(x, y)=(0, 1)$ does not satisfy equation $(1)$ or $(2)$ .

Case 2: If $2y=0\implies y=0$

Then $(2x)^3=(x+0+1)^3\implies 2x=x+1\implies x=1$ Upon checking, $(x, y)=(1, 0)$ is indeed a solution to both $(1)$ and $(2)$ .

Case 3: If $x+y+1=0$

Then $(2x)^3=-(2y)^3\implies x=-y$ But then $x+y+1=x-x+1=1\neq0$ . Oh noes! A contradiction to our original assumption!

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Therefore, $(x, y)=(1, 0)$ is the only integer solution.