Integer of form 6k + 5 is of form 3k + 2 but not Conversely
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Theorem
Let $n \in \Z$ be an integer of the form:
- $n = 6 k + 5$
where $k \in \Z$.
Then $n$ can also be expressed in the form:
- $n = 3 k + 2$
for some other $k \in \Z$.
However it is not necessarily the case that if $n$ can be expressed in the form:
- $n = 3 k + 2$
then it can also be expressed in the form:
- $n = 6 k + 5$
Proof
\(\ds n\) | \(=\) | \(\ds 6 k + 5\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \paren {2 k} + 3 + 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3 \paren {2 k + 1} + 2\) | for some $2 k + 1 \in \Z$ |
Replacing $2 k + 1$ with $k$ gives the result.
$\Box$
Now consider $n = 8$.
We have that:
- $8 = 3 \times 2 + 2$
and so can be expressed in the form:
- $n = 3 k + 2$
However:
- $8 = 6 \times 1 + 2$
and is in the form:
- $n = 6 k + 2$
and so cannot be expressed in the form:
- $n = 6 k + 5$
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $2$: Divisibility Theory in the Integers: $2.1$ The Division Algorithm: Problems $2.1$: $2$