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Find the number of positive integers n for which $n^{2}+96$ is a perfect square.
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$n^{2} + 96 = x^{2}.$
$x^{2} - n^{2} = 96.$
$(x-n)(x+n) = 96.$

Since, $x$ and $n$ both should be positive integer. $(x-n)$ and $(x+n)$ will be divisors of $96.$

By observation, $(x-n)$ should be smaller than $(x+n)$ because $x$ and $n$ are positive integers.
$(x-n) = k1 \rightarrow x= n+k1$

($x+n)= k2 \rightarrow n+k1+n = k2 \rightarrow 2n+k1 = k2 \rightarrow 2n = k2-k1 \rightarrow n=\left(\frac{k2-k1}{2}\right).$

As we have seen from above, $n = \left ( \frac{k2-k1}{2} \right )$

Therefore, for n to be a positive integer,$k2-k1$ should be even. That is, both should be odd or both should be even.

There are 6 pairs of divisors when multiplied becomes $96 = (1,96), (2,48), (3,32), (4,24), (6,16), (8,12) .$

Therfore, there are only 4 such possibilties. Phew!

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No of positive integer is 1

value is +2

Now to solve it

n2+96 = x2

or, x2 - n2 = 96

or, (x+n)(x-n) =96

or,(x+n)=12

(x-n) = 8

x=10

n=2

Answer:

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