Mathematica The Sum of All Possible Values for a and b in a Pythagorean Equation

[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?

 

[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.

 

[arildno]

Zurtex was the one going into the trap..

 

[Data]

a, b and c are positive prime numbers

Be careful!

 

[arildno]

Be careful!

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

 

[Data]

such evil people here!


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.

 

[Zurtex]

Zurtex was the one going into the trap..

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

 

[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.

 

[Werg22]

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

 

[Werg22]

such evil people here!


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.

Not at all.

 

[Berislav]

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

 

[Werg22]

I'm terribly sorry I meant whole numbers.

 

[Data]

haha, that changes the whole question

 

[arildno]

I'm terribly sorry I meant whole numbers.

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.

 

[Werg22]

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

 

[Zurtex]

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

 

[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

 

[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

 

[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...