9555=3∗5∗7^2∗13
So 9555^2=3^2∗5^2∗7^4∗13^2
This has (2+1)(2+1)(4+1)(2+1) or 135 factors out of which one is 9555 and 67 are below 9555 and 67 are above 9555
(a + c) > 9555 and (a-c) < 9555 is a set of solution
So number of solutions 67
For a-c we need to take3x∗5y∗7z∗13m (x,y,z,m to be chosen based on limit for example x between 0 and 3 such that a +c > 9555) and for a-c it is 9555^2/(a+c)
So 9555^2=3^2∗5^2∗7^4∗13^2
This has (2+1)(2+1)(4+1)(2+1) or 135 factors out of which one is 9555 and 67 are below 9555 and 67 are above 9555
(a + c) > 9555 and (a-c) < 9555 is a set of solution
So number of solutions 67
For a-c we need to take3x∗5y∗7z∗13m (x,y,z,m to be chosen based on limit for example x between 0 and 3 such that a +c > 9555) and for a-c it is 9555^2/(a+c)
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