Prove that every natural number $n>15$ there exist natural numbers $x,y\geqslant1$ which solve the equation. Prove that every natural number $n>15$  exist Natural numbers $x,y\geqslant1$ which solve the equation $3x+5y=n$.
so i try induction.
base case is for $n=16$.
so $\gcd(5,3)=1$, after Euclidean algorithm i found:
$3(32-5t)+5(-16+3t)=16$ so i found that $32-5t>1$c for $x$ and $-16+3t>1$ for   $y$  and for $t=6$ $x=2$ and $y=2$ .
now suppose its takes place for $n$.
how i show that for $n+1$?
if there is more elegant way i would love to see.
 A: For $n=16$, it is easy: $16=2\times3+2\times5$.
Now, let $n\in\mathbb{N}\setminus\{1,2,\ldots,14\}$ and suppose that there are natural numbers $x$ and $y$ such that $n=3x+5y$. Then:


*

*if $y>1$,\begin{align}n+1&=3x+5y+1\\&=3(x+2)+5(y-1)\end{align}and $x+2,y-1\in\mathbb{N}$.

*Otherwise, $n+1=3x+5+1=3x+6$, for some natural number $x$ and so$$n+1=3(x-3)+5\times3.$$Note the $x-3\in\mathbb N$, since\begin{align}n+1\geqslant16&\iff3x+6\geqslant16\\&\iff3x\geqslant10\end{align}and therefore, since $x\in\mathbb N$, $x\geqslant 4$.

A: If $x,y$ is a solution, you have $3x+5y=n=n(-3\cdot3+2\cdot5)$ and therefore
$$3(x+3n)= 5(2n-y)$$
As $3,5$ are coprime it exists $k \in \mathbb Z$ such that $x=5k-3n$ and $y= 2n-3k$.
$x >1$means $5k> 1+3n$ and $y>1$ implies $3k<2n-1$. 
Which is equivalent to 
$$3+9n < 15k < 10n-5 \tag{1}$$
The difference of $10n-5$ and $9n+3$ is equal to $n-8$. For $n-8>15$, i.e $n>23$, finding $k$ satisfying $(1)$ is obviously always possible. Remain the cases $n=16, \dots , 23$ that can be evaluated one by one.
Example of $n=5847$
With this analysis, it is easy to find the solution. We have $$9\cdot 5847+3=52626<15\cdot 3509 =52635 <58465.$$
Hence a solution is $x=5\cdot 3509-3\cdot 5847=4$ and $y=2\cdot 5847-3\cdot 3509=1167$.
We can verify that $3 \cdot 4+ 5\cdot 1167=5847$.
