I have $xy$, can i get $x+y$? I was trying to solve one of my university math problem and it basically came down to this question. If i get an answer or some hints i would be able to solve my original problem.
It sounds pretty simple or dum, but i could not figure it out.
If i have $x$ and $y$ as natural numbers and i know $xy$ value, can i get possible $x+y$ values?
Or maybe you guys know some relations between multiplication and addition results?
I know that there is a rule that:
$$\frac{x+y}2 \ge \sqrt{xy}$$
I was thinking about it for a while now.
Any tips and hints would be thankful.
EDIT: $x$ and $y$ are unknown.
Thanks.
 A: Let's say the product $P=xy=12$, and you know that $x$ and $y$ are positive integers.
It's straightforward to find all possible values of $x+y$, and in this case you can write them all out easily.  (I'll assume without loss of generality that $x \leq y$.)
$$x=1, y=12: x+y = 13 \\ x=2, y=6: x+y = 8 \\x = 3,y=4: x+y = 7$$
So it's one of those three values, but without more information (like $y-x=4$) you can't know which one for sure.
A: Yes, to a degree that you can classify these.
Let $xy = \prod_{i}p_i^{a_i}$ be the prime factorization of $xy$.
The $x = \prod_{i}p_i^{b_i}$ where $0 \le b_i \le a_i$ and $y = \prod_{i}p_i^{a_i - b_i}$ and $x + y = \prod_{i}p_i^{b_i}+ \prod_{i}p_i^{a_i - b_i}$.  Due to symmetry there are $\lceil \frac{\prod (a_i + 1)}2 \rceil$ solutions.
So for example if $xy = 7  =7^1$ then there is one solution $x+y = 1+7 = 8$.
If $xy = 12 = 2^2*3$ then there are $3$ solutions: $x + y = \{1+12, 2+6, 3+4\}=\{13,8,7\}$
If $xy = 2^6*3^4*5^2=129600$ will have $53$ solutions:
$1+129600= 129601, 2+64800=64802, 4+32400=32404, 8+16200=16208, 16 + 8100=8116, 32+4050=4082, 64 + 2025=2089, 3+43200=43203, 6+21600=21606, .....$.
More reasonable: $xy = 48 = 2^4*3$ will have $5*2/2=5$ solutions:
$1+48 = 49, 2+24=26, 4+12 = 16, 8+6 = 14, 16+3=19$
