re: Perfect square sum problem

Quote:

Originally Posted by

**emilhp** Extend the fraction 163/101 with a number such that the resulting fraction will have a numerator and a denominator whose sum are square numbers. Please provide the smallest of such number.

let $\displaystyle \frac{163}{101} = \frac{a}{c}$

$\displaystyle \frac{a}{c} = \frac{ab}{cb}$

sum of the numerator and denominator ...

$\displaystyle ab+cb = b(a+c)$

for the sum $\displaystyle b(a+c)$ to be a perfect square, $\displaystyle 264b = (2^3 \cdot 3 \cdot 11 \cdot b)$ would have to be a perfect square ...

re: Perfect square sum problem

Isn't extending a fraction just multiply it by a number? If that is the case, wouldn't the answer be

$\displaystyle (\frac{161}{101})(\frac{161}{101})=(\frac{161^2}{1 01^2}) or \frac{25921}{10201}$

Re: Perfect square sum problem

Quote:

Originally Posted by

**skeeter** let $\displaystyle \frac{163}{101} = \frac{a}{c}$

$\displaystyle \frac{a}{c} = \frac{ab}{cb}$

sum of the numerator and denominator ...

$\displaystyle ab+cb = b(a+c)$

for the sum $\displaystyle b(a+c)$ to be a perfect square, $\displaystyle 264b = (2^3 \cdot 3 \cdot 11 \cdot b)$ would have to be a perfect square ...

How would you find b?

Re: Perfect square sum problem

Quote:

Originally Posted by

**emilhp** How would you find b?

come on, I have to leave something for you to think about ...

Re: Perfect square sum problem

Quote:

Originally Posted by

**mdhafn** Isn't extending a fraction just multiply it by a number? If that is the case, wouldn't the answer be

$\displaystyle (\frac{161}{101})(\frac{161}{101})=(\frac{161^2}{1 01^2}) or \frac{25921}{10201}$

If the OP wanted to square the original fraction, this would be the case.