Is it possible to find coprime natural "a" & "b" in the equation
a^(2) + b^(2) = c^(2)
When "c" (a natural even number) is given?
2006-11-25 17:05:47 · 5 answers · asked by rajesh bhowmick 2 in Science & Mathematics ➔ Mathematics
I know how to find out "c" & "a" when "b" is given & I am always getting "c" as odd when trying for natural coprime solutions of "a" & "b".
2006-11-25 17:24:06 · update #1
Yes. When 'c' is given, find the value of c^2. Then, find out the possible numbers, whose sum = that number. Find out the root of all these numbers, till you arrive a perfect square root( if possible ).
Eg: when given, c = 10, find c^2. Here, it is 100.
therefore, a few possibilities are : 50 + 50 ( none are perfect squares)
25 + 75 ( only 1 of them is a perfect square)
36 + 64 ( both are perfect squares)
Therefore, the answers being looked for are 6 and 8.
2006-11-26 23:18:26 · answer #1 · answered by Enlightened 2 · 0⤊ 0⤋
If c is even, a and b must be both odd or both even. They cannot be coprime if they are both even.
2006-11-28 03:15:59 · answer #2 · answered by Seshagiri 3 · 0⤊ 0⤋
If c is even, a and b must be even, and therefore cannot be coprime.
2006-11-26 02:09:17 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
You're looking for Pythagorean Triples:
3,4,5
6,7,10
5,12,13
7,24,(5^2)^2
the above from the second link...
http://mathworld.wolfram.com/PythagoreanTriple.html
http://www.geocities.com/fredlb37/node1.html
2006-11-26 01:13:24 · answer #4 · answered by modulo_function 7 · 0⤊ 0⤋
yes these are just pythagorian triplets. you should express a & b in terms of c
2006-11-26 11:40:26 · answer #5 · answered by gjmb1960 7 · 0⤊ 0⤋