document.write( "Question 1127550: The general formula for the Pythagorean triples is a=m^2-n^2, b=2mn and c=m^2+n^2 for integers m and n. How would this formula be used to find all the number of primitive Pythagorean triples containing 24 as one number? \n" ); document.write( "

Algebra.Com's Answer #743978 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( "2mn is always even.

\n" ); document.write( "If m and n are both even, then both m^2-n^2 and m^2+n^2 are even, and the triple would not be primitive.
\n" ); document.write( "If m and n are both odd, then again both m^2-n^2 and m^2+n^2 are even, and the triple is again not primitive.

\n" ); document.write( "So for a primitive triple, we need the following:
\n" ); document.write( "(1) m and n are one even and one odd
\n" ); document.write( "(2) m > n
\n" ); document.write( "(3) 2mn = 24 --> mn = 12

\n" ); document.write( "There are 3 pairs of integers m and n that meet conditions (2) and (3):
\n" ); document.write( "(m,n) = (12,1); (6,2); (4,3).

\n" ); document.write( "The second pair does not satisfy condition (1), so there are exactly 2 primitive Pythagorean Triples containing the number 24:

\n" ); document.write( "m = 12, n = 1 --> the triple is (143, 24, 145)

\n" ); document.write( "and

\n" ); document.write( "m = 4, n = 3 --> the triple is (7, 24, 25) \n" ); document.write( "

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