# Let 5a+12b and 12a+5b be the side lengths of a right-angled triangle and 13a+kb be the hypotenuse, where a, b and k are positive integers. How do you find the smallest possible value of k and the smallest values of a and b for that k?

Feb 25, 2018

$k = 10$, $a = 69$, $b = 20$

#### Explanation:

By Pythagoras' theorem, we have:

${\left(13 a + k b\right)}^{2} = {\left(5 a + 12 b\right)}^{2} + {\left(12 a + 5 b\right)}^{2}$

That is:

$169 {a}^{2} + 26 k a b + {k}^{2} {b}^{2} = 25 {a}^{2} + 120 a b + 144 {b}^{2} + 144 {a}^{2} + 120 a b + 25 {b}^{2}$

$\textcolor{w h i t e}{169 {a}^{2} + 26 k a b + {k}^{2} {b}^{2}} = 169 {a}^{2} + 240 a b + 169 {b}^{2}$

Subtract the left hand side from both ends to find:

$0 = \left(240 - 26 k\right) a b + \left(169 - {k}^{2}\right) {b}^{2}$

$\textcolor{w h i t e}{0} = b \left(\left(240 - 26 k\right) a + \left(169 - {k}^{2}\right) b\right)$

Since $b > 0$ we require:

$\left(240 - 26 k\right) a + \left(169 - {k}^{2}\right) b = 0$

Then since $a , b > 0$ we require $\left(240 - 26 k\right)$ and $\left(169 - {k}^{2}\right)$ to have opposite signs.

When $k \in \left[1 , 9\right]$ both $240 - 26 k$ and $169 - {k}^{2}$ are positive.

When $k \in \left[10 , 12\right]$ we find $240 - 26 k < 0$ and $169 - {k}^{2} > 0$ as required.

So the minimum possible value of $k$ is $10$.

Then:

$- 20 a + 69 b = 0$

Then since $20$ and $69$ have no common factor larger than $1$, the minimum values of $a$ and $b$ are $69$ and $20$ respectively.