Each positive integer is painted with some color. we know that $a+b$ and $ab$ have the same colour if $a>1, b>1$.

Question

Each positive integer is painted with some color. It is known that for all pairs $a, b$ of integers greater than $1$ the numbers $a+b$ and $ab$ have the same color. Prove that all numbers greater than $4$ are of the same color.

I've found that if $a+b=x$ $ab=k$ are the same colour then $(a-1)(b+1)=q$ is the same colour too because $a-1+b+1=x$ And like that if some $a+b=x$ there are many $ab$ that have same colour I think I've to write with rows and come back , go forward again and prove that all numbers are the same colour.

Answer 1

Are you familiar with the concept of induction?

Bear with me. Suppose $N=2K$ is a somewhat large even number. Then $N$ is the same color as $2+K$ and $2+K < N=2K$ if $K > 2$ or in other words if $N=2K> 4$

If $N=2K + 1$ is a somewhat large odd number then $N=(2K-2)+3$ so $N$ is the same color as $3(2K-2)=6(K-1)$. so $N$ is the same as $6+(K-1)=K+5$. And $K+5 < N=2K+1$ if $K>4$ or in other words if $N=2K + 1 > 9$.

So if we can show that the colors for $4,5,6,7,8,9$ are all the same color we are done as we can reduce from any larger number down to those.

$2+3=5;2*3=6$

$6=2+4=3+3; 2*4=8;3*3=9$

$12 = 2*6 = 3*4$ and $2+6=8$ and $3+4=7$.

so, yes, $4,5,6,7,8,9$ (and $12$) are the same color.

So that's it, we're done.

Answer 2

Let $a\sim b$ mean that numbers $a$ and $b$ have the same color. So for any $n>4$ we have $n\sim 2(n-2)$, because $2+(n-2)=n$. Similarly, $2(n-2)\sim 4(n-3)$ because $2+2(n-3) = 2(n-2)$. Hence $$n\sim 4(n-3)\sim 4+(n-3)=n+1$$ So by induction, all integers greater than 4 have the same color.