# Perfect square sum problem

• Dec 8th 2011, 07:02 AM
emilhp
Problem
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.

I've had some problems with this task, hope that you can help me.(Happy)
• Dec 8th 2011, 09:54 AM
skeeter
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 $\frac{163}{101} = \frac{a}{c}$

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

sum of the numerator and denominator ...

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

for the sum $b(a+c)$ to be a perfect square, $264b = (2^3 \cdot 3 \cdot 11 \cdot b)$ would have to be a perfect square ...
• Dec 8th 2011, 09:56 AM
mdhafn
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

$(\frac{161}{101})(\frac{161}{101})=(\frac{161^2}{1 01^2}) or \frac{25921}{10201}$
• Dec 8th 2011, 11:29 AM
emilhp
Re: Perfect square sum problem
Quote:

Originally Posted by skeeter
let $\frac{163}{101} = \frac{a}{c}$

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

sum of the numerator and denominator ...

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

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

How would you find b?
• Dec 8th 2011, 12:23 PM
skeeter
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 ...
• Dec 8th 2011, 01:07 PM
Quacky
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

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

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