<s>[INST] Proof that $n = 3k + 5l$ for $n > 7$ 
Show that for every n greater than $7$, there are non-negative integers $k$ and $l$ such that $$n = 3k+ 5l.$$

So induction seems like a possibility.
$n = 3k + 5l$ and so $n + 1 = 3k + 5l + 1$. 
What can be done in such a case? Can I get a hint?
 [/INST] By strong induction:  We can write
\begin{align*}
8 & = 1 \cdot 3 + 1 \cdot 5\\
9 & = 3 \cdot 3 + 0 \cdot 5\\
10 & = 0 \cdot 3 + 2 \cdot 5
\end{align*}
Let $n \geq 10$.  Assume that we can write each integer $m$ such that $8 \leq m \leq n$ in the form $m = 3k + 5l$ for some non-negative integers $k, l$.  We wish to show that $n + 1$ can also be written in this form.  Since $n + 1 \geq 10 + 1 = 11$, $n + 1 - 3 = n - 2 \geq 8$, by the induction hypothesis, there exist non-negative integers $k'$ and $l'$ such that $n - 2 = 3k' + 5l'$.  Therefore, $$n + 1 = n - 2 + 3 = 3k' + 5l' + 3 = 3(k' + 1) + 5l'$$  Hence, each integer $n > 7$ can be expressed in the form $n = 3k + 5l$, where $k$ and $l$ are non-negative integers.     
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