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I need to find $a, b$ positive integers that verify: $$a + b = 1271$$ $$\mathrm{lcm}(a, b) = 330\gcd(a, b)$$

I don't know how to start thinking those kind of exercises.

user26857
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nnov
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  • What are a* and b*? – TMM Apr 13 '17 at 02:42
  • It's just a notation for: $$ \frac{a}{gcd(a,b)} = a^$$ $$\frac{b}{gcd(a,b)} = b^$$

    so we can simplify: $$\text{lcm}(a, b) = \frac{ab}{gcd(a,b)} = 330.\gcd(a, b)$$ $$a^b = 330.\gcd(a, b)$$ $$\frac{a^b}{gcd(a,b)} = 330$$ $$a^b^ = 330$$ Now that you point it, I remember that notation from university, but I don't know if it's official as I've looked around and haven't seen it anywhere.

     – nnov Apr 13 '17 at 03:00

1 Answers1

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Hint: First of all, $\text{gcd}(a,b)=\text{gcd}(a,1271-a)$. Now suppose $d\mid a$ and $d\mid (1271-a)$, then there exist number $k,l\in \mathbb{Z}$ such that $a=kd$ and $1271-a=ld$. Hence $$1271-a=1271-kd=ld.$$ Thus $$31\cdot 41=1271=(k+l)d.$$ What does this tell you about $k+l$ and $d$? What if you suppose that $d=\text{gcd}(a,b)$? Also, recall that $\text{lcm}(a,b)\text{gcd}(a,b)=ab$, this might help.

Edit: Suppose $d=\text{gcd}(a,b)$. Since $d\mid 31\cdot 41$, either $d=1,31,41$ or $1271$. Now we can work by cases:

Case I: If $d=1$, then $\text{lcm}(a,b)=330=2^2\cdot 3\cdot 5\cdot 11=ab$ and $a+b=1271$. Now there are only finitely many possibilities for $a$ and $b$ since they should be divisors of $330$. You can check by hand.

Case II: If $d=31$, then $\text{lcm}(a,b)=\frac{ab}{31}=2^2\cdot 3\cdot 5\cdot 11\cdot 31$. So $ab\mid 2^2\cdot 3\cdot 5\cdot 11\cdot 31^2$ again there are finitely many cases.

The other cases can be dealt with similarly. With some more thought you can severly reduce the amount of work you need to do (There is one more trick to make the calculation quite easy). But since you blatantly copied most of my answer into your question, I'll leave the rest up to you.

At least this method helps you to reduce to finitely many cases which in principle can be checked.

Mathematician 42
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  • Thank you for the answer. I've been stuck with this exercise for a few hours now, thinking I would find some properties that would be the key element to solve the problem. Didn't think I just had to do a case analysis. I didn't intend to copy your work, it's my first day here, and people downgraded the post. Looked around why would people downgrade it and it said somewhere that it might look like I wasn't doing any research, etc. - so I thought I had to do some work there. – nnov Apr 13 '17 at 03:08
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    Ok no problem. In principle the equation $a+b=1271$ only has finitely many solutions in the positive integers, so you could work from here. Working as above you severely reduce the amount of cases. It's still a lot, but many cases are easy to deal with. – Mathematician 42 Apr 13 '17 at 08:52