# Can anyone solve this?

1. Sep 21, 2005

### Werg22

This question goes:

a, b and c are positive integrers.

What is the sum of all possible values of a and b between 0 and 100 if

a^(-2) + b^(-2)=c^(-2)

2. Sep 21, 2005

### gerben

$$c = (\sqrt{a}+\sqrt{b})^2$$

$$a \in [0, 100] \rightarrow \sqrt{a} \in [0, 10]$$

so the sum of all values of a is:

$$\sum_{x=1}^{10}{x^2}$$

the same for b.

3. Sep 22, 2005

### VietDao29

I don't understand...
$$a ^ {-2} = \frac{1}{a ^ 2}$$, and that's not $$a ^ {-2} = \sqrt{a}$$
Viet Dao,

Last edited: Sep 22, 2005
4. Sep 22, 2005

### Werg22

I've solved it...

1/c^2=1/b^2 + 1/a^2
=(a^2+ b^2)/(ab)^2
c^2=ab^2/(a^2+ b^2)
c=ab/(a^2+ b^2)^1/2

Since a^2+ b^2 is an integrer, it's root is either another integrer or irrational. Thus in order for c to be an integrer, (a^2+ b^2)^1/2 must be an integrer.
Listing possible result;

3^2 + 4^2 = 5^2
5^2 + 12^2 = 13^2
20^2 + 21^2 = 29^2
....

Then considering the first possibility, a=4x and b=3x

c=12x^2/5x
=12x/5

We conclude that x must be a factor of 5. Since a=4x, a is multiple of 20. Adding up all multiples of 20 between 0 and 100

20 + 40 + 60 + 80 + 100=300

Now for b, b=3x

15 + 30 + 45 + 60 + 75 + 90=315

Now the second possibility,

c=60x/13

a=5x, b=12x

x must be a multiple of 13

5(13)=65, and 12(13)>100.

The next possibility, we knoe that a=20x and b=21x, and x has to be a factor of 29. Since 21(29)>20(29)>100, then there is no further solution.

300 + 315 + 65=680.

So the awnser is 680.