System of equations with different exponents

Question

For integers $x$ and $y$, $$x^3+29y=2018$$ $$29x+y^3=1348$$ Could someone please tell how to start/solve this problem without a calculator. I tried to find $x$ in terms of $y$ and then substitute, but the numbers became too ugly. Also, please don't give away the answer. Thanks in advance!

Answer 1

It would be easy if $x$ and $y$ are integers.

\begin{align*} x^3-y^3+29y-29x&=670\\ (x-y)[(x-y)^2+3xy-29]&=670 \end{align*}

$670=2\times 5\times 67$.

$(x-y,(x-y)^2+3xy-29)=(1,670)$, $(2,335)$, $(5,134)$, $(10,67)$, $(67,10)$, $(134,5)$, $(335,2)$ or $(670,1)$.

Most cases can be rejected. Only possible $(x,y)$ are $(12,10)$.

Non-integer solutions are much more difficult to find.

Answer 2

The only integral solution is $(x,y)=(12,10)$. Actually, this is the only real solution. Using resultants, I obtain (with a calculator, I admit), that either $x=12$, or

$$ x^8 + 12x^7 + 144x^6 - 4326x^5 - 51912x^4 - 622944x^3 + 4741644x^2 + 56899728x + 682089455=0, $$ which has $8$ complex, non-real solutions.

Answer 3

For positive integer solutions, it is enough to note that the equations imply

$$ x^3 \equiv 17 \pmod{29} \\ y^3 \equiv 14 \pmod{29} $$

By brute force, checking $\,n^3 \bmod 29\,$ for $\,-14 \le n \le 14\,$ finds that the unique solutions are:

$$ x \equiv 12 \pmod{29} \\ y \equiv 10 \pmod{29} $$

It turns out that $x=12,y=10$ satisfy the equations, and since $x \lt \sqrt[3]{2018} \simeq 12.64\,$, $y \lt \sqrt[3]{1348}\simeq 11.05$ this is the only solution in positive integers.