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a and b are consecutive positive integers. So,suppos b=a+1
D=a^2+b^2+c^2=a^2+(a+1)^2+(a)*(a+1)=3a^2+3a+1=3(a^2+a)+1
Since a is a positive number and (a^2+a) is always an even number (Doesn't matter whether a is an odd number or an even number) and so 3(a^2+a) will also be an even number. Therefore, 3(a^2+a)+1 is an odd number.
So Square root of D will always be an odd positive number
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