I'm trying to solve the following equation:


I got:


But I can't go any further. I am looking for a solution in integers.

I need some hints to proceed, please!

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    $\begingroup$ I found one solution by inspection: if $x-y=1$, you have $x^2+y^2=61$, which has $x=5,y=6$ as a solution. $\endgroup$ – Brian M. Scott May 2 '12 at 7:13
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    $\begingroup$ @Gigili: Given that there are two unknowns, and only one equation, how many answers do you suppose there would be? $\endgroup$ – Joel Cornett May 2 '12 at 7:19
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    $\begingroup$ @Joel: There may be only finitely many solutions in integers. $\endgroup$ – Brian M. Scott May 2 '12 at 7:20
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    $\begingroup$ If $x-y\geq 1$, it follows that$(x-y)(x^2+xy+y^2)=(x-y)xy+(x-y)(x^2+y^2) \geq xy + (x^2+y^2)$ and this allows us to bound the search space within the integers. Looking for such bounds, however crude, is one technique for dealing with this kind of question. $\endgroup$ – Mark Bennet May 2 '12 at 7:22
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    $\begingroup$ @Gigili Adding to Mark Bennet suggestion, once you get some bounds you can restrict your sample space by arguments based on parity/modular arguments. For instance, in the current problem a simple parity argument gives us that both $x$ and $y$ cannot be even. Similarly, if both are odd, then one has to be $1 \bmod 4$ and the other has to be $3 \bmod 4$. Which with some further arguments give us that both cannot be odd. Hence, you can always conclude that one has to be even and other has to be odd. $\endgroup$ – user17762 May 2 '12 at 7:47

Put $x+y=:u$, $x-y=:v$; then $u$ and $v$ are integers of the same parity. In the new variables the equation becomes $${1\over8}(6u^2v+2v^3)={1\over4}(u^2-v^2)+61\ ,$$ which can be written as $$27 u^2+(3v+2)^2={6584\over3 v-1}\ .$$ It follows that $v>1$ and that $3v-1$ must divide $6584=8\cdot 823$; whence $v\in\{1,3,549,2195\}$. $549$ and $2195$ are obviously too large, and $v:=3$ leeds to $u^2=26$ which has no integer solutions. It remains $v=1$ which leads to $u^2=121$ or $u=\pm11$. Therefore we only have the two solutions $(x_1,y_1):=(6,5)$ and $(x_2,y_2):=(-5,-6)$.


I believe the only answers in the integers are $(6,5)$ and $(-5,-6)$.

If $x$ and $y$ are integers, then we can let $y=x-k$, where $k$ is an integer. Then the equation can be written as a quadratic in $x$: $$k(x^2+x(x-k)+(x-k)^2)-x(x-k)-61 = 0$$ The solutions to this are: $$x = \frac{k(3k-1) \pm\sqrt{(1-3k)(k^3+k^2-244)}}{2(3k-1)}$$ The term under the square root is only positive for $k=1,2,3,4,5$, and checking the results for each of these shows that only $k=1$ results in integer solutions (listed above).

  • $\begingroup$ Thank you for your answer, how did you get that the term under the square root is only positive for $K=1, \dots, 5$? $\endgroup$ – Gigili May 2 '12 at 9:26
  • $\begingroup$ It is a quartic polynomial in $k$ with leading coefficient $-1$. $\endgroup$ – copper.hat May 2 '12 at 15:11
  • $\begingroup$ Oops, I meant $-3$. The $k^3+k^2-244$ term is increasing in $k$ and is positive when $k \geq 6$, negative when $k \leq 5$. Similarly, the $1-3k$ term is positive for $k\leq 0$, negative for $k\geq 1$. The only integers for which the product is non-negative are $1,...,5$. $\endgroup$ – copper.hat May 2 '12 at 17:36

If we restrict ourselves to positive solutions, we can do it by brute force. In that case clearly $x>y>0$.

If $x-y=1$, we have $x^2+y^2=61$, which has the unique solution $x=6,y=5$.

If $x-y=2$, we have $2(x^2+xy+y^2)=xy+61$, or $2(x^2+y^2)+xy=61$; clearly $x$ and $y$ must both be odd, with $x^2+y^2<30$. The only candidate is $x=3,y=1$, which clearly doesn't work.

If $x-y=3$, we have $3(x^2+y^2)+2xy=61$; here $x$ and $y$ must have opposite parity, with $x^2+y^2<20$, so $x=4,y=1$ is the only candidate, and it doesn't (quite) work.

If $x-y=4$, we have $4(x^2+y^2)+3xy=61$, so $x$ and $y$ must be odd, and $x^2+y^2<15$; there are no solutions in this case.

Suppose that $x-y\ge 4$. Then $(x-y)(x^2+xy+y^2)\ge 4(5^2+5\cdot1+1^2)>61$, so there are no solutions.

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    $\begingroup$ Dear Brian, if we draw $x^3-y^3=z$ and $xy+61=z$ just for examining the functions, they intersect eachother infinitely. $\endgroup$ – mrs May 2 '12 at 9:00

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