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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.

José Carlos Santos
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arthur
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2 Answers2

1

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 $x\in\mathbb N$, $x\geqslant 4$.
José Carlos Santos
  • 427,504
1

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 $3 \cdot 4+ 5\cdot 1167=5847$.

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