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I know that product of two consecutive integers must be even, but not too sure how it helps in proving this.

yociyoci
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  • Do you know when a product of two consecutive numbers is divisible by 6? – Sam Weatherhog Nov 11 '15 at 22:08
  • when at least one of the consecutive integer is divisible by 3 – yociyoci Nov 11 '15 at 22:11
  • That's right! So if you don't want that to happen, what's the remainder when you divide your first number by 3? – Sam Weatherhog Nov 11 '15 at 22:12
  • the odd number would have a remainder of 1, and the even number would have a remainder of 2 – yociyoci Nov 11 '15 at 22:14
  • We could operate crudely. If we have $2$ consecutive integers, the smallest is of form $6k$ or $6k+1$ or $\dots$ or $6k+5$. In all but the case $6k+1$ and $6k+4$ the product is divisible by $6$. So only two cases to look at. – André Nicolas Nov 11 '15 at 22:14
  • You are right @yociyoci but the number with remainder 1 can be even (e.g. 10 and 11). But the first number must have remainder 1 mod 3. You can use this to prove the rest – Sam Weatherhog Nov 11 '15 at 22:17

2 Answers2

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Let $n$ and $n+1$ be the consecutive integers. The first thing to note is that if $n\equiv 0$ (mod $3$) or $n+1\equiv 0$ (mod $3$) then the product will be divisible by $6$ (since one of them must also be even). So the only possibility is that $n\equiv 1$ (mod $3$). Working mod $9$ now we have the following possibilities:

$$ n\equiv 1, 4, 7 \text{ (mod }9\text{)} $$

For each of these we have:

$$ n(n+1)\equiv 2,2,2 \text{ (mod }9\text{)} $$

So in all cases, the product is of the form $9t+2$ for some $t\in \mathbb{Z}$.

Sam Weatherhog
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The product of two consecutive numbers is always even, so we just have to deal with the factor of 3. Two consecutive numbers having no factor of 3 between them must mean that they sit between multiples of 3, i.e. the numbers are $3k+1$ and $3k+2$ for some value $k$. Then their product is $\left(3k+1\right)\left(3k+2\right) = 9k^2 + 3k + 6k + 2 = 9k^2 + 9k + 2 = 9\left(k^2 + k\right) + 2$ as required.

ConMan
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