# Little mathematical problem

1. Apr 30, 2005

### Werg22

This is not for help or anything, I just to see your level of mathematical cleverness.

Knowing that a, b and c are positive whole numbers;

$$a^{-2} + b^{-2} = c^{-2}$$

What is the sum of all possible values for a or b from 0 to 100?

Last edited: Apr 30, 2005
2. Apr 30, 2005

### Zurtex

Well that's the same as:

$$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$$

$$\frac{a^2 + b^2}{(ab)^2} = \frac{1}{c^2}$$

Therefore a2 + b2 must be a divisor of (ab)2 and ab=c.

The rest is a simple matter of writing a program to work the possibilities out under this and adding them together.

Last edited: Apr 30, 2005
3. Apr 30, 2005

### arildno

Zurtex was the one going into the trap..

4. Apr 30, 2005

### Data

Be careful!

5. Apr 30, 2005

### arildno

You are a gentleman, Data!
I was hoping for a hurried response along the lines:
"Trap?! What trap?"

6. Apr 30, 2005

### Data

such evil people here! :rofl:

Now, if the poster were going to be really devious, he'd say that the answer is

$$\sum_{ 0 \leq p_i \leq 100} p_i, \ p_i \ \mbox{prime},$$

with the claim that $a, \ b, \ c$ don't appear in the equation he gave at all - he meant

$$\hat{a} + \hat{b} - 4= \hat{c} -2,$$

and he also specified that you choose either $a$ or $b$ to add the values for, so it's just the sum of all primes between 0 and 100.

Last edited: Apr 30, 2005
7. Apr 30, 2005

### Zurtex

Haha, I only opened the door, I didn't go down it :tongue:

8. Apr 30, 2005

### Werg22

Zurtex you are right on track, but not quite. You can solve it, if you have the proper thinking. One hint; square root the equation.

9. Apr 30, 2005

### Werg22

$$\frac{\sqrt{(a^2+b^2)}} {ab} = \frac{1} {c}$$

Last edited: Apr 30, 2005
10. Apr 30, 2005

### Werg22

Not at all.

11. Apr 30, 2005

### Berislav

Is it, by any chance, zero?
ab=c, which goes against the definition of a prime.

12. Apr 30, 2005

### Werg22

I'm terribly sorry I meant whole numbers.

13. Apr 30, 2005

### Data

haha, that changes the whole question :tongue:

14. Apr 30, 2005

### arildno

How sad; I thought you had made a really good trap Zurtex fell into, even though he strenuously asserts he only opened the (trap-) door a bit.
What he forgot, is the influence of gravity:
If you step upon a trap-door, opening it, however little, gravity pulls you down to the bottom.

15. Apr 30, 2005

### Werg22

It would have been impossible with prime numbers, so there is no trap :p

16. Apr 30, 2005

### Zurtex

pfft, whatever, clearly didn't go as far as Werg, anyway I didn't make a single mistake in my answer

17. Apr 30, 2005

### Werg22

Now with the equation try awnsering it!

Now awnser? I'll give another hint. Inverse the equation

$$\frac{ab} {\sqrt{a^{2} + {b^2}}} = c$$

If no one can awnser by the end of the day I will awnser it myself

Last edited: May 1, 2005
18. May 1, 2005

### Werg22

Ok here how to solve it.

First, if a, b and c are whole numbers then $${\sqrt{a^{2}$$ has to be a certain number that will devid ab in a whole number. Now the result of $${\sqrt{a^{2}$$ is either whole or decimal... not quite. Phytagorus with whole numbers NEVER give decimal numbers that are perfect

19. May 1, 2005

### Werg22

Ok here how to solve it.

First, if a, b and c are whole numbers then $${\sqrt{a^{2} + b^{2}}$$ has to be a certain number that will divide ab in a whole number. Now the result of $${\sqrt{a^{2} + b^{2}}$$ is either whole or decimal... not quite. No product of whole numbers is devidable by their Pythagoaras result into another whole number. So, now you know that $${\sqrt{a^{2} + b^{2}}$$ has to be a perfect Pythagoras. But witch numbers when multiplied give a number that is dividable by those two numbers Pythagoras? Only 5 factors. Witch numbers give 5 factors perfect Pythagoras? Factors of 3 and 4. So a and b

have to be:

Factors of 5
And factors of 3 or 4

$$3(5)=15$$

$$4(5)=20$$

Every perfect Pythagoras with factors of 3 and 4 have common factor witch we will name x.

$$a=15x$$
$$b=20x$$

$$\frac{15x(20x)} {\sqrt{(15x)^{2} + {(20x)^2}}} = c$$

$$\frac{300 x^2} {\sqrt{225 x^2 + 400 x^2}}} = c$$

$$\frac{300 x^2} {\sqrt{x^2 625}}} = c$$

$$\frac{300 x^2} {25x}= c$$

$$12x= c$$

All the possible values for a and be would be

$$15(x+1+1+...)+20(x+1+1...)$$

The maximum value for x in the case of 15 is 6 since 15*6=90 and 15*7=105
The maximum value for x in the case of 20 is 5 since 20*5=100 and 20*6=120

The sum for all numbers to 0 to 6 is

$$\frac{6^2 + 6} {2}=21$$

and 5

$$\frac{5^2 + 5} {2}=15$$

SO

$$15(21) + 20(15)=615$$

But there is ONE exception. That is 65.

$$\frac{65(156)} {\sqrt{65^{2} + {156^2}}} = 169$$

$$615+65=680$$

680 is the sum of all possible values for a and b. I must admit this problem disturbed a while and it took me almost a week to solve it! I had to rediscover all of this. I guess a quite clever mathematician would have been able to solve it right away...

Last edited: May 1, 2005