Original problem

Show that there are exactly 16 pairs of integers $(x,y)$ such that $11x+8y+17=xy$.

My work

From case by case analysis I come to know that the equation will hold if and only if $x$ is odd and $y$ is even.

Also I found that $(8-x)|(11x+17)$ and $(11-y)|(8y+17)$

This is all what I have found.

Please see that my work are right or not.

This is a new kind of question which I have encountered so please help me is solving this problem.

  • $\begingroup$ All your conclusions are correct and the best way I know is described in my answer ($x$ is odd and $y$ is even is a correct conclusion and $8-x$ divides $11x+17$ and $11-y$ divides$8y+17$ are valid consequences) $\endgroup$ – Elaqqad May 2 '15 at 14:31
  • $\begingroup$ @Elaqqad Sir this was a 10 marks problem of last year paper of an institute entrance exam. I guess that my conclusions only award me 1 or 2 marks. (math.stackexchange.com/questions/1261971/…) Please look at this problem. This one of those 10 marks problem. $\endgroup$ – Singh May 2 '15 at 14:35

Given $a,b,c$ three integers, the idea when you have an equation of the form: $$xy=ax+by+c$$ to solve if for unknown integers $x,y$ is to do the following factorization: $$(x-b)(y-a)=c+ab$$

and hence the number of solutions is the number of divisors of $c+ab$.

| cite | improve this answer | |
  • $\begingroup$ After a search I found (mathwarehouse.com/answered-questions/factors/…) please look this. It is showing that the divisors of 105 is 8. Please explain. $\endgroup$ – Singh May 2 '15 at 14:38
  • $\begingroup$ what if you count the negative ones also you will have $16$ divsors and so there are $16$ solutions in total $\endgroup$ – Elaqqad May 2 '15 at 14:40
  • $\begingroup$ I missed that. Thank you Sir. One more help, out of 10 how much you will give for my conclusions. $\endgroup$ – Singh May 2 '15 at 14:49

This type of diophantine equation is solvable by a generalization of completing the square. Namely, completing a square generalizes to completing a product as follows:

$$\begin{eqnarray} &&axy + bx + cy\, =\, d,\ \ a\ne 0\\ \overset{\times\,a}\iff\, &&\!\! (ax+c)(ay+b)\, =\, ad+bc\end{eqnarray}\qquad\qquad$$

Here we deduce $\ xy - 11x - 8y = 17 \iff (x-8)(y-11) =\, \ldots$

So by uniqueness of prime factorizations the problem reduces to counting the divisors of $\,\ldots$

| cite | improve this answer | |

16 ordered pairs indeed.

$$\begin{align} 11x+8y+17&=xy \\ xy-11x-8y&=17 \\ xy-11x-8y+\color{red}{88}&=17+\color{red}{88} \\ (x-8)(y-11)&=105=3 \cdot 5 \cdot 7=d_1 \cdot d_2 \end{align}$$


$$\begin{cases} x-8=d_1\\ y-11=d_2 \end{cases}$$

Implies that $$\begin{cases} x=8+d_1\\ y=11+d_2 \end{cases} \text{such that} \ \ (d_1,d_2) \in \lbrace \text{divisor-pairs of 105} \rbrace$$ For any divisor-pair $(d_1,d_2)$ that produces one solution, $(d_2,d_1)$ will produce another.

Additionally, for any $(d_1,d_2)$ that produces a solution, $(-d_1,-d_2)$ will produce another as well.

So for every divisor-pair, $(d_1,d_2)$, there will be four total solutions:
$$\begin{align} (d_1,d_2) &\to (d_1,d_2) \\ &\to (d_2,d_1) \\ &\to (-d_1,-d_2) \\ &\to (-d_2,-d_1) \end{align}$$

This is due to a lack of symmetry in the expressions for $x$ and $y$ (for comparison, this situation is different).

The problem therefore reduces to finding exactly $4$ divisor-pairs of $105$.
Since $$105=3^{\color{red}{1}} \cdot 5^{\color{red}{1}} \cdot 7^{\color{red}{1}} \implies 105 \ \text{has} \ (1+{\color{red}{1}})(1+{\color{red}{1}})(1+{\color{red}{1}})=8 \ \ \text{divisors},$$

there are indeed four divisor-pairs: $$(d_1,d_2) \in \bigg{\{} (1,105),(3,35),(5,21),(7,15) \bigg{\}}$$

As an example, setting $d_1=1$ and $d_2=105$

$$\begin{align} (1,105) &\to (1,105) && \implies (x,y)=(9,116)\\ &\to (105,1) && \implies (x,y)=(113,12)\\ &\to (-1,-105) && \implies (x,y)=(7,-94)\\ &\to (-105,-1) && \implies (x,y)=(-97,10) \end{align}$$

That's four solutions from one divisor-pair alone.

| cite | improve this answer | |

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.