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