I can't test them all -2

Number Theory Level 4

Find the sum of all values of positive integer $n$ such that $n^2+96$ is a perfect square .

The answer is 40.

1 solution

Peter Macgregor
Apr 15, 2016

$n^2+96=p^2$

$\implies p^2-n^2=96$

$\implies (p+n)(p-n)=96$

The numbers are small enough to allow a case by case treatment.

Notice that the first term $(p+n)$ is greater than the second one $(p-n)$ . This allows us to list all the possible cases. For each row the entries in the third and fourth columns come from solving the first and second columns as simultaneous equations for $n$ and $p$ . I have omitted non-integer results because they are of no interest to us! -

$\begin{pmatrix}(p+n)&(p-n)&p&n\\96&1\\48&2&25&23\\32&3\\24&4&14&10\\16&6&11&5\\12&8&10&2\\&&\text{total}&\boxed{40}\end{pmatrix}$

Great solution. Upvoted! Bonus: What should be done if the number is large and case by case treatment is cumbersome?

Rishik Jain - 5 years, 2 months ago

Letting $(p + n) = a$ and $(p - n) = b$ , we want to find integers $a, b$ such that $ab = 96$ . Note that $a$ and $b$ must both be even. Thus, we can eliminate the work of checking the factors $(1, 96)$ and $(3, 32)$ . This is why we end up with non-integer results for them.

Zach Abueg - 3 years, 10 months ago

I can't test them all -2

The answer is 40.

1 solution

0 pending reports