# Thread: Proof by Contradiction: Sum of Squares of 3 consecutive integers

1. ## Proof by Contradiction: Sum of Squares of 3 consecutive integers

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.

2. 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.
You have $\displaystyle 3 n^2 + 6n +6 -12a=0$
so
$\displaystyle n^2 +2n +2 -6a = 0$.

In order for this quadratic equation to have an integer root, the discriminant
$\displaystyle 4 - 4(2-6a) = 2^2 (6a -1)$
must be a perfect square, so
$\displaystyle 6a-1$ must be a perfect square, i.e.
$\displaystyle 6a -1 = x^2$.
Looking at this modulo 6,
$\displaystyle x^2 \equiv -1 \pmod{6}$.

So try all the possible values of x mod 6: 0, 1, 2, 3, 4, 5. Do any of them work?