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Given $x$ and $y$ are multiples of $2$ satisfying $$x^2 - y^2 = 27234702932$$ Find the number of solutions to $x$ and $y$.

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4 Answers 4

up vote 6 down vote accepted

As $27234702932=2^2\cdot181\cdot37616993$ where the last two factors are primes

If $x=2X,y=2Y$

$$X^2-Y^2=181\cdot37616993$$

If $X^2-Y^2=p\cdot q$ where $p,q$ are primes,

the possible cases for $X+Y,X-Y$ are $\pm pq, \pm 1$ and $\pm p,\pm q $

For example, if $X+Y=1,X-Y=pq$

The number of positive factors of $p\cdot q$ is $(1+1)\cdot(1+1)=4$

So, there should be $4\cdot2 =8$ solutions in integers.

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If $X^2-Y^2=\prod_{1\le r\le n}p_i^{r_i},$ the number of factors combinations will be $2\cdot \prod_{1\le r\le n}(r_i+1)$ –  lab bhattacharjee May 13 '13 at 12:04
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HINT $(X-Y)(X+Y)=X^2-Y^2$

(then factor)

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thanks for the reply . Can you make your point less hazy pls ?? –  Charan Pai May 13 '13 at 11:51
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First factor $27234702932=2^2\times181\times37616993$. Since both are even, so is $x+y$ and $x-y$. Each solution is uniquely determined by $x+y$ and $x-y$.
Hence the number of solutions is equal to the number of solutions to $ab=27234702932$ where both $a,b$ are even.

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Factorize the number and use $x^2-y^2=(x-y)(x+y)$?

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