SOLUTION: Prove that if the two larger numbers in a pythagorean triple differ by 1 , then their sum is the square of the smaller number (example 5, 12 ,13 )

Question 1199421: Prove that if the two larger numbers in a pythagorean triple differ by 1 , then their sum is the square of the smaller number (example 5, 12 ,13 )
Answer by ikleyn(52787) (Show Source): You can put this solution on YOUR website!
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Prove that if the two larger numbers in a pythagorean triple differ by 1,
then their sum is the square of the smaller number (example 5, 12 ,13 )
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Let "a", "b" and "c" be a pythagorean triple;

"b" abd "c" be two larger numbers (with "c" be the LARGEST) and 

    c - b = 1    (1)   (differ by 1).


Then 

    a^2 + b^2 = c^2,

hence,

    a^2 = c^2 - b^2 = (c-b)*(c+b).

Sunstitute here 1 instead of c-b, based on (1), and you will get

    a^2 = b + c.


This equality is what should be proved.


At this point, the proof and the solution of the problem is complete.

Solved.

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