Originally Posted by

**Harry1W** I have been working through D. L. Johnson's "Elements of Logic via Numbers and Sets" and have come across this problem, amongst others, which has me stumped.

"Give a proof by contradiction of the statement:

$\displaystyle P $: the sum of the squares of three consecutive integers cannot leave remainder $\displaystyle -1 $ on division by $\displaystyle 12 $."

Therefore, I have suggested that we assume (for contradiction) ¬P ('not' P), so

$\displaystyle \frac{n^2 + (n+1)^2 + (n+2)^2}{12} = a - \frac{1}{12} $,

where $\displaystyle n $ is the lowest integer in the consecutive series and $\displaystyle a $ is an integer.

Then,

$\displaystyle \begin{array}{ccc}

\frac{3n^2 +6n + 5 - 1}{12} & = & a \\

\frac{3n^2 + 6n +4}{12} & = & a

\end{array} $.

I think for the purposes of our proof, I wish to show that $\displaystyle a $ cannot be an integer (so providing the necessary contradiction). I'm not sure, however, how to do this, and so whether this former assertion is correct. Any suggestions much appreciated.