If $a^2+b^2+c^2=D$ and a and b are consecutive positive integers and $ ab = c $ then prove that $ D^{1/2} $ is an odd integer

Question

Here is my approach: since a and b are consecutive integers there are two possibilities $a = b+1$ and $a = b-1$ substituting this into the first equation we get $ 2 b^2 + 2 b + c^2 + 1 = D $ after this I do not know what to do. help in any form would be appreciated.

Answer 1

Without loss of generality, let $a = b+1$. Then $$a^2+b^2+c^2 = (b+1)^2+b^2+b^2(b+1)^2 = (b^2+b+1)^2$$ Now, consider that whether $b$ is even or odd, $b^2+b$ will be even (as it is either an even number plus an even number, or an odd number plus an odd number), so $b^2+b+1$ is odd.

Answer 2

here is how you can do it: you have already done a portion of it and got $ 2 b^2 + 2 b + c^2 + 1 = D $ now use $ab=c$ and substitute $a=b-1$ or $a=b+1$ (they both lead to the same thing) to get $b(b+1)=c$ or $b(b-1)=c$ plug this value of c to get $c+c^2+1=D$, $(c+1)^2=D$, so $D^{1/2} = c+1$ and since the product of consecutive integers ($a$ and $b$) is always even it means $c$ is even so $c+1$ is odd