Can a and b be odd and still have a solution to "c"
Geometry
Level
2
In a set of pythogorean triplets satisfying the equation the equation a^2 + b^2 = c ^ 2 there cannot be an integer solution to c to when both a and b are odd. Is this True or False ?.
False
True
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1 solution
If a and b are odd then they are of the form a = 2p +1 and b = 2k+ 1 .
Writing a^2 + b^ 2 in this form , a^2 + b^2 = 4p^2 + 4p + 1 + 4k^2 + 4k + 1 or a^2 + b^2 = 4p^2 + 4k^2 + 4p + 4k + 2. This is definitely an even number. If a solution to a^2 + b^ 2 = c^2 exists then c^2 should be even as (a^2 + b^ 2 ) is even when a and b are odd. If c is even , then c is of the form 2 j so c^2 is 4 j^2 where j is an integer > 0
Therefore ( a^2 + b^2) mod 4 = 2 ,whereas (c^2 )mod 4 = 0. This cannot be possible and hence no solution exists for c when a and b are odd integers and when a,b and c are pythogorean triplets.