 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Quadratic and Linear (Posted on 2016-08-03) Find all possible triplets (A,B,C) of positive integers satisfying this system of simultaneous equations:
• A2 + B2 - C2 = A+B-C, and:
• A+B+C = 82

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 2 of 2 | Rearrange the second equation into C=82-A-B and substitute into the first equation to yield A^2 + B^2 - (82 - A - B)^2 = A + B - (82 - A - B)

This equation simplifies into A*B - 81A - 81B + 3321 = 0.  Then some factoring makes (A-81) * (B-81) = 3240.

3240 has 40 factorizations into integers, including negative factorizations.  Only two of those factorizations result in positive (A,B,C).  3240 = -54 * -60 generates the solution (27,21,34) and 3240 = -72 * -45 generates the solution (9,36,37).

 Posted by Brian Smith on 2017-07-02 22:44:38 Please log in:

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