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Suppose that $a$ and $b$ are positive integers such that $ab$ divides $a + b$.

Prove that $a = b$, then prove that $a$ is either $1$ or $2$.

I was thinking using the theorem:

If $n|a$ and $n|b$,then $n|(ax+by)$ for any $x,y\in \mathbb Z$.

Then I can derive that $ab$ divides $a$ and $b$. I could not go further. However, this doesn't seem like can help me solve the question I had.

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closed as off-topic by The Phenotype, ahulpke, kimchi lover, Nosrati, Parcly Taxel Feb 5 '18 at 6:51

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  • $\begingroup$ If you can provide some context for the question, ideally including your own thoughts on it, then you'll avoid having your question voted down and possibly closed. This site isn't very friendly towards plain questions without effort being shown on the part of the asker. $\endgroup$ – G Tony Jacobs Feb 5 '18 at 0:34
  • $\begingroup$ I was thinking using the theorem If n|a and n|b,then n|(ax+by) for any x,y∈Z.Then I can derive that ab divides a and divides b. I could not go further. However, this doesn't seem like can help me solve the question I had. $\endgroup$ – Xiang Feb 5 '18 at 0:39
  • $\begingroup$ I think I could go if a divides b and b divides a then a equals b. $\endgroup$ – Xiang Feb 5 '18 at 0:44
  • $\begingroup$ @Xiang Simple things first... The product of two positive integers is usually larger than their sum. When is it not so? $\endgroup$ – dxiv Feb 5 '18 at 0:44
  • $\begingroup$ that's right, therefore a is either 1 or 2. but how can you prove this? $\endgroup$ – Xiang Feb 5 '18 at 0:47
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We have $$ab \mid a+b \implies ab \mid b( a+b) \implies ab \mid b( a+b) - ab \implies ab \mid b^2 \implies a \mid b$$ In a similar way we can show $$b \mid a$$ and so $$a=b$$

If we substitute $b$ by $a$ in $ab \mid a+b $ we get

$$a^2 \mid 2a \implies a \mid 2 \implies a \in \{1,2\}.$$ Both $(a,b)\in\{(1,1), (2,2)\}$ are solutons.

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$a$ divides $b \iff \exists k\in\mathbb{Z}$ such that $b=ka$.

Now, $ab$ divides $a+b\iff \exists K\in\mathbb{Z}$ s.t $a+b=Kab$.

If $a=b\Rightarrow a+a=Kaa\Rightarrow 2a=Ka^2$, remember that $a\neq0$. Then

$2=Ka$ where you have two ways.

What integer values should K take to satisfy the equation?

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abx=(a+b), where x is greater than or equal to 1 (x cannot be 0)

x = 1/a + 1/b

This implies that x must be 1 or 2. (2 is the greatest value possible for sum of two reciprocals of natural numbers)

so if x=1, then 1/a and 1/b must both be 1/2 and a=b=2. so if x=2 then 1/a and 1/b must both be 1 and a=b=1

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