Katie S.

asked • 09/01/14

finding integers

find three positive integers a,b and c such that a+b+c=1000 and a2+b2=c2

Inactive Tutor

At the outset I thank Katie for asking the question and Fellow Tutors Surendra K. and Phillip R. for their inputs. Katie's question induced me to read some fascinating material on the internet about “Pythagorean Triples”. Though I came some 50+ years ago from the country that gave the world Number-Theorists like Brahmagupta and Ramanujan, Number Theory is not my specialization. It is Statistics.
 
However, i like doing research and I think I have solved the generalization of Katie's problem completely, obtaining a characterization. The proof also extends the idea of Phillip R.

My appeal to my Fellow Tutors is whether or not they would be interested in sparing the time to read my proof, in the spirit of camaraderie.  They are welcome to criticize and find mistakes. I shall be forever indebted to them.
 
Please post a comment. If there is interest, I shall give the proof in several parts of posted comments.
 
Thank you all.
 
Dr. G.
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09/07/14

3 Answers By Expert Tutors

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Inactive Tutor answered • 09/01/14

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a, b, and must be sides of a right triangle with c the hypotenuse. And 1000 is a multiple of the sum. one combination comes to mind is 8, 15, 17. The sum is 40 so we multiply each number by 25. 
So we have a = 200, b= 375, and c = 425. 
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Inactive Tutor

My heartfelt appreciation for Phillip R.!  He found at least. one solution.
 
This also suggests the following:
 
From the website mentioned in my answer, find all those triplets of positive integers (A, B,C) whose SUM IS A FACTOR OF 1000. Call that factor R.  (in Phillip R.'s solution R= 25).
 
Then a solution is a = RA, b = RB, c = RC.
 
Of course, this is not a com[lete set of solutions!
 
Dattaprabhakar(Dr. G.)
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09/01/14

Inactive Tutor answered • 09/01/14

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Katie:
 
I was thinking about this problem while running an errand.  I have made some changes.
 
As Tutor Surendra K. has pointed out, there are many possible solutions.
 
From a+b+c = 1000, we get c = 1000 - a - b.
 
From the second equation, we get a2 + b2 = (1000 - a - b)2 .
 
So
 
a2 + b2 = 10002 + a2 + b2 - 2000a - 2000b + 2ab.
 
That is,  2000a + 2000b - 2ab =10002.
 
That is,  1000 a + 1000 b - ab = 500000
 
If you choose intelligently a value for b, you can solve for a and hope that a is a positive integer!
 
With b = 100, say  (Will it work?)
 
1000 a + 10000 - 100 a = 500000
 
 
900 a = 490000  NO, a is not an integer.
 
All I can suggest right now is what is called a "computer intensive" method.  Write a computer program as follows:
 
Step 1: Choose positive integers (a, b, c) such that a2 + b2 = c2.
Step 2: Check whether or not a + b + c = 1000. 
          Step 2a: If YES, you have found one solution. 
          Step 2b: If NO, choose a different triplet of positive integers (a, b, c) satisfying the condition of Step 1.
Step 3: Go to Step 2.
 
Incidentally a partial list of triplets satisfying the condition in Step 1, is available on the website
 
http://en.wikipedia.org/wiki/Pythagorean_theorem
 
Katie, let me think on this problem some more.  If I hit upon an analytical solution I will get back to you.  Wish me luck!  In the meantime, if you find something, please post a comment.
 
Dattaprabhakar (Dr. G.)
 
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