Can You Find a Number n with Rational Square Roots for n-7, n, and n+7?

[musicgold]

Hi,

I am struggling with this puzzle from a book.

Puzzle : Can you find a number n such that, the numbers n-7, n, and n+7 have rational square roots (can be expressed as integers or fractions)?
According to the book one of the solutions is n =113569 /14400

This is what I have done so far:

Let p, q, r be the square roots of n-7, n, and n+7, respectively.

(n-7) * n * (n+7) = p^2 * q^2 * r^2

n^3 -49n = p^2 * q^2 * r^2

As I have 4 unknowns and only one equation, I do not know how to proceed from here. What should I do?

Thanks.


Cross posted at:
http://mathforum.org/kb/thread.jspa?threadID=2370848

 

[Vargo]

I would suggest setting n=(p/q)^2 with p,q relatively prime.

Then if (p^2/q^2) - 7 = a^2/b^2, what can you say about b? Then you work the other condition on n+7 similarly and then you will end up with 2 equations, both of which look quite similar to the equation for Pythagorean triples. I didn't finish the problem, so I can't guarantee this line of reasoning.