Need to solve below question:

Find all solutions in integers $x$ and $y$ of the equation->> $xy + 5x -8y = 79$

Any hint is welcome.


It's $$xy+5x-8y-40=39$$ or $$(y+5)(x-8)=39,$$ and solve a number of systems: $$x-8=-39$$ and $$y+5=-1...$$

For $$x-8\in\{-39,-13,-3,-1,1,3,13,39\}$$ you'll get all integer solutions.

  • $\begingroup$ I am not clear about the values (a,b) chosen for the rhs of the two equations (x - 8 = a; y + 5 = b). I feel the values should be such that a.b = 39. Then, these values can be derived from the set { 1, 13, 39, 3}. One value can be chosen, and the other derived based on the quotient = 39/ (earlier chosen value). Still, both will be from the same set. $\endgroup$ – jiten Oct 25 '17 at 18:19
  • 1
    $\begingroup$ @jiten Yes, of course! $x-8\in\{-39,-13,-3,-1,1,3,13,39\}$ and you'll get all integer solutions. $\endgroup$ – Michael Rozenberg Oct 25 '17 at 18:22
  • 1
    $\begingroup$ @jiten proceeding this way is available: $$(y+5)(x-8)=39=d_1 \cdot d_2$$, $$\implies$$ $$\begin{cases}x=8+d_1 \\ y=-5+d_2\end{cases}$$ such that $$(d_1,d_2) \in \{(1,39),(39,1), (3,13),(13,3),(-1,-39), (-39,-1),(-3,-13),(-13,-3)\}$$So there should be 8 solutions total. It's worth noting that this situation lacks symmetry, so switching $d_1$ and $d_2$ will give different solutions, besides that, I don't know if all this adds any worthwhile clarity or not but just a small suggestion. $\endgroup$ – MaximusFastidiousIrreverence Oct 26 '17 at 14:50
  • $\begingroup$ @AmateurMathGuy Your comment regarding symmetry adds a lot for looking the solution by an alternate view. I never knew that symmetry would be there in the solution sets obtained; i.e. had symmetry been there, 4 solutions would have been there. Can you please point a problem with such symmetry. $\endgroup$ – jiten Oct 31 '17 at 16:38
  • $\begingroup$ @jiten I wanted to express that the expressions arrived at will give different solutions for $x$ and $y $ if you were to switch the values of $d_1$ and $d_2$. Suppose instead we had that $$x=8 + d_1 + d_2$$. Then since we have the symmetric expression $ d_1 + d_2$, switching the values won't change the result $\endgroup$ – MaximusFastidiousIrreverence Oct 31 '17 at 17:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.