Dynamic Diophantine!

Level 2

Find the number of ordered pairs $(x,y)\in\mathbb{Z}^2$ that satisfies the following equation:

$xy+9(x+y)=2006$

The answer is 4.

2 solutions

Jubayer Nirjhor
Jan 6, 2014

$\begin{aligned} &~ &xy+9(x+y)&=&2006 \\ \\ &\Longrightarrow ~~~ &9x+xy+9y+81~~~&=~&2006+81 \\ \\ &\Longrightarrow ~~~ &(x+9)(y+9)~~~&=~&2087 ~~~~~~~[2087~\text{is a prime}] \\ \\ &~&~&=&1\times 2087 \\ \\ &~&~&=&2087\times 1 \\ \\ &~&~&=&(-1)\times (-2087) \\ \\ &~&~&=&(-2087)\times (-1) \\ \\ \end{aligned}$

$\therefore ~~~ (x,y)\in \left\{(-8,2078),(2078,-8),(-10,-2096),(-2096,-10)\right\}$

$\text{Total:}~~~\fbox{4}~\text{pairs!}$

Hmm. But your solution might not be smart enough if the number had many divisors, and you may not be always be able to use Simon's favorite factoring tactic(SFFT) like this. My approach is the general one to handle cases like this. :)

not real - 7 years, 4 months ago

It's wiser to use divisor theorem. $2087$ is a prime so I didn't find it important to explain further. :)

Jubayer Nirjhor - 7 years, 4 months ago

Not Real
Jan 18, 2014

We can re-write the equation as $x(y+9)=2006-9y$ This gives us $y+9|2006-9y$ but since $y+9|9y+81$ , subtracting we have $y+9|9y+81+2006-9y=2087$ . So every divisor of $2087$ associates with a corresponding $y$ , regardless of negative or positive. Since $2087$ is a prime number, $2087$ has a total of 4 divisors(negative and positive).

Dynamic Diophantine!

The answer is 4.

2 solutions

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