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

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Discussion Overview

The discussion revolves around the mathematical exploration of the equation \( a^{-2} + b^{-2} = c^{-2} \), where \( a \), \( b \), and \( c \) are positive whole numbers. Participants are investigating the sum of all possible values for \( a \) or \( b \) within the range of 0 to 100, while also considering the implications of prime numbers and the nature of the equation.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant reformulates the original equation into a more manageable form, suggesting that \( a^2 + b^2 \) must be a divisor of \( (ab)^2 \) and that \( ab = c \).
  • Another participant introduces the idea that if \( a \), \( b \), and \( c \) are prime numbers, it complicates the situation, implying a potential trap in the problem's setup.
  • Several participants engage in playful banter about the existence of traps and the cleverness of the problem, indicating a light-hearted atmosphere.
  • A participant suggests that the sum of all prime numbers between 0 and 100 could be a hidden answer, although this is met with skepticism.
  • Another participant emphasizes the need to consider whole numbers instead of primes, which shifts the focus of the problem.
  • One participant provides a detailed breakdown of how to approach solving the equation, discussing the relationship between \( a \), \( b \), and \( c \) and the conditions under which they yield whole numbers.
  • A later reply presents a comprehensive calculation leading to a proposed sum of 680 for all possible values of \( a \) and \( b \), while also noting an exception for the value 65.

Areas of Agreement / Disagreement

Participants express a range of views, with no clear consensus on the approach to the problem or the validity of the proposed solutions. Disagreements arise regarding the interpretation of the equation and the implications of using prime numbers versus whole numbers.

Contextual Notes

Participants highlight the complexity of the problem, noting that the definitions of numbers (whole vs. prime) significantly impact the discussion. There are also unresolved mathematical steps and assumptions that participants rely on throughout the conversation.

Werg22
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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?
 
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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.
 
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Zurtex was the one going into the trap..
 
a, b and c are positive prime numbers

Be careful!
 
Data said:
Be careful!
You are a gentleman, Data!
I was hoping for a hurried response along the lines:
"Trap?! What trap?" :biggrin:
 
such evil people here! :smile:


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.
 
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arildno said:
Zurtex was the one going into the trap..
Haha, I only opened the door, I didn't go down it :-p
 
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.
 
\frac{\sqrt{(a^2+b^2)}} {ab} = \frac{1} {c}
 
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  • #10
Data said:
such evil people here! :smile:


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.
 
  • #11
Is it, by any chance, zero?
ab=c, which goes against the definition of a prime.
 
  • #12
I'm terribly sorry I meant whole numbers.
 
  • #13
haha, that changes the whole question :-p
 
  • #14
Werg22 said:
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.
 
  • #15
It would have been impossible with prime numbers, so there is no trap :p
 
  • #16
:biggrin: pfft, whatever, clearly didn't go as far as Werg, anyway I didn't make a single mistake in my answer :rolleyes:
 
  • #17
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 :frown:
 
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  • #18
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
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...
 
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