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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.

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  • $\begingroup$ 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
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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 $c+ab$.

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  • $\begingroup$ 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
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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 $\,\ldots$

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16 ordered pairs indeed.

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

Setting

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

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.

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