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I need to prove that:

If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$

So far I used what's given so I have:

$ax+by=1$ (I wrote the gcd of a and b as a linear combination)

and

$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)

where do I go from here?

Martin Sleziak
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Lil
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5 Answers5

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Assume this is not true.

Let gcd$(a+b,ab)=m>1$. Then there exists a prime number $p$ which divides $m$.

If $p\mid a+b$ and $p\mid ab$, then $p$ divides $a$ or $b$.

Assume that $p\mid a$. But then $p\mid a+b$ implies that $p\mid b$, and hence $p\mid\,$gcd$(a,b)$, which is a contradiction.

Note. We have used the fact that: If $p$ is a prime and $p$ divides $ab$ then $p$ divides $a$ or $b$.

Yiorgos S. Smyrlis
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  • How do you justify "If $p\mid a+b$ and $p\mid ab$, then $p$ divides $a$ or $b$."? It's not immediately obvious how that follows. – Jeff Snider Mar 02 '14 at 17:19
  • can this question be proven using a direct proof? I'm a little confused by your proof using contradiction – Lil Mar 02 '14 at 17:19
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    @JeffSnider: If a prime $p$ divides a product $ab$, then $p$ divides at least on of $a$ and $b$. – Yiorgos S. Smyrlis Mar 02 '14 at 17:21
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    I might just want to add why it is a contradiction. By hypothesis we were given $\gcd(a,b) = 1$, and we showed that $ p \mid \gcd(a,b) $. By substitution this implies $p \mid 1 $. But this is not possible since prime numbers $p$ are integers greater than or equal to 2. – john Jan 17 '21 at 07:30
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Below is a proof that is messier but more primitive than Yiorgos' proof. By Bezout's Identity, we have that there exists $x,y \in \mathbb{Z}$ such that

$$ax + by = 1.$$

Squaring both sides, we see that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = 1.$$

But notice that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2).$$

And hence, those same $x,y$ as above give

$$ ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2) = 1.$$

So it must be that $\gcd{(a+b,ab)} = 1$.

Gamma Function
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  • alright it makes sense up to this part. ab(2xy−x^2−y^2)+(a+b)(ax^2+by^2).Can you explain how you factored a^2x^2+2abxy+b^2y^2? – Lil Mar 02 '14 at 17:47
  • I understand a+b(ax^2+by^2)but for the other part I would have ab(2xy) – Lil Mar 02 '14 at 17:49
  • I may have made a mistake in my algebra. Let me check it over. – Gamma Function Mar 02 '14 at 17:57
  • no I just checked you are correct. I just wouldn't have thought to factor x^2-y^2 but if you don't add those terms it doesn't come back to the original – Lil Mar 02 '14 at 17:58
  • Ok. One can look for $u$ and $v$ such that $(ab)u + (a+b)v = 1$. Right off the bat, you can group the $2abxy$ with the first term, and follow your nose from there. If you have already seen the relevant theorems (ie Euclid's Lemma), Yiorgos' proof is almost certainly prefered. If you haven't seen Euclid's Lemma, you may want to look into its proof as its a pure gem. – Gamma Function Mar 02 '14 at 18:03
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I like pushing the limits of the Euclidean algorithm. Let's do our best to elimiante $b$'s:

$$\begin{align} \gcd(a+b, ab) &= \gcd(a+b, ab - a (a+b)) \\&= \gcd(a+b, -aa) \\&= \gcd(a+b, a^2) \end{align}$$

Since we also have $\gcd(a+b, a) = 1$, we can infer that $\gcd(a+b, a^2) = 1$.

2

We're given $\gcd(a,b) = 1$, and from Bézout's identity we have $\gcd(a,b) = 1 \iff \exists x,y: ax + by = 1$ for integer $x$ and $y$.

$1 = ax + by = ax + bx - bx + by = (a + b)x + b(y - x)$, so $\gcd(a+b,b) = 1$. Likewise, $\gcd(a+b,a)=1$ because $(a + b)y + a(x - y) = 1$

$$ \begin{align} 1 &= ((a + b)x + b(y - x))((a + b)y + a(x -y)) \\&= (a+b)(a+b)xy + (a+b)a(x-y) + (a+b)b(y-x)y + ab(y-x)(x-y) \\&= (a+b)[(a+b)xy + a(x-y) + b(y-x)] + ab[(y-x)(x-y)] \end{align}$$

So $\gcd(a+b,ab) = 1$.

NovaDenizen
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Proof broken into lemmas:

Lemma 1: If $n$ is rel. prime to $a$ and $n$ is rel. prime to $b$, then $n$ is rel. prime to $ab$.

Proof: multiply together the Bezout identities for the two hypotheses.

Lemma 2: If $a$ and $b$ are rel. prime, then both are rel. prime to $a+b$.

Proof: Its enough to prove if $a$ and $b$ are rel. prime then $b$ is rel. prime to $a+b$. We have the Bezout identity: $a x + by = 1$. Now add and subtract $bx$ on the left.

Conclusion: We are given $a$ and $b$ rel. prime. By the 2nd lemma, $a+b$ is rel. prime to both $a$ and $b$. By the first lemma, $a+b$ is rel. prime to $ab$.

fred goodman
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