Rearranging the equation we get: \[c=\frac{17\cdot 2^3}{a^2}+\frac{17 \cdot 3^2 \cdot 5}{b^2}\]
Because the LHS is an integer, the RHS must also be an integer. First, let's address the case in which both fractions in the RHS are integers. Then \(a=1, 2\) and \(b=1, 3\), giving us \(4\) solutions so far.

Now note that when \(a > 11\) and \(b > 27\), the RHS is less than one, and thus cannot be a positive integer. You could easily exhaust all 297 possibilities, although I'm sure there is a better way.
–
Bob Krueger
·
3 years, 9 months ago

## Comments

Sort by:

TopNewestRearranging the equation we get: \[c=\frac{17\cdot 2^3}{a^2}+\frac{17 \cdot 3^2 \cdot 5}{b^2}\] Because the LHS is an integer, the RHS must also be an integer. First, let's address the case in which both fractions in the RHS are integers. Then \(a=1, 2\) and \(b=1, 3\), giving us \(4\) solutions so far.

Now note that when \(a > 11\) and \(b > 27\), the RHS is less than one, and thus cannot be a positive integer. You could easily exhaust all 297 possibilities, although I'm sure there is a better way. – Bob Krueger · 3 years, 9 months ago

Log in to reply