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How many number of ordered pairs of positive integer $(a,d)$ satisfies:

$$\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=2012$$

This question is too complicated. Like I tried to multiply by the common denominator, but it's not going anywhere since it's very complex. I tried to put it on wolframalpha and my ti89 they don't return me with anything.

Any clues?

Thanks!

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  • $\begingroup$ It's not really too complicated. Expand out and you should find LHS equals $4 a + 6 d$. $\endgroup$ – Andreas Caranti Apr 27 '13 at 11:01
  • $\begingroup$ To help you obtain @Andreas' expression, first simplify the denominators; now what is $\frac1{p/q}$? $\endgroup$ – Lord_Farin Apr 27 '13 at 11:02
  • $\begingroup$ $\frac{q}{p}$ right? $\endgroup$ – C Connie Apr 27 '13 at 11:02
  • $\begingroup$ BTW, I did it by hand, but checked using GAP, so it's strange that your computer algebra systems did not help. $\endgroup$ – Andreas Caranti Apr 27 '13 at 11:03
  • $\begingroup$ @CConnie Indeed. (How convenient of me and Andreas to both delete our comment :).) $\endgroup$ – Lord_Farin Apr 27 '13 at 11:04
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The equation simplifies as

\begin{align} &\frac{1}{\frac{1}{a+2d}-\frac{1}{a+3d}}-\frac{1}{\frac{1}{a}-\frac{1}{a+d}}=\frac{1}{\frac{d}{(a+2d)(a+3d)}}-\frac{1}{\frac{d}{a(a+d)}}\\ &=\frac{(a+2d)(a+3d)}{d}-\frac{a(a+d)}{d}\\&=\frac{a^2+5ad+6d^2-a^2-ad}{d}\\&=\frac{4ad+6d^2}{d}\\&=4a+6d\\&=2012 \end{align}

Thus, $2a+3d=1006$. Then, we can take this equation by modulo $2$, getting $d\equiv 0\ (mod \ 2)$, thus $d$ is even.

Now let $d=2x$, thus $x\in\mathbb{N}$. Taking the equation in modulo $3$, we get $2a\equiv 1\ (mod \ 3)$, thus $a\equiv 2\ (mod\ 3)$. Now, let $a=3y-1$, thus $y\in\mathbb{N}$.

By doing substitution, we get:

$$2(3y-1)+3(2x)=1006\implies 6y-2+6x=1006\implies x+y=168$$

Thus, there're $\boxed{167}$ ordered pairs where $a,d\in\mathbb{N}$.

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The equation is equivalent (a > 0; d > 0) to

4a + 6d = 2012 or

2a + 3d = 1006

the solution is

a = 503 - 3n

d = 2n

Since a and d should be positive

503 - 2n > 0

2n > 0

So we can derive that n = 1..167 and finally we have 167 solutions:

500, 2

497, 4

...

2, 334

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This is another approach not involving modulus. $2a+3d=1006$. $2a$ will always be even no matter what a is, and you need 2 even numbers to add up to an even number, so the term $3d$ also has to be even. This can only be accomplished if $d$ itself is even. Now since $a$ can be $1, 2, 3, ...$ the term $2a$ can be any even number. So now solutions can be something like 1004+2, 1002+4, etc. (not all of these will be solutions of course). Now for any even $d$ you will have a solution for $a$ because $a$ can be any positive integer, so let's find the maximum number of solutions for $d$. We know that $3d<1006$, so $d<$ $or$ $=335$ Remember that $d$ has to be even, so how many even numbers are there between $1$ and $335$? Well half of them, which is $167$ which is your answer.

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