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Here is Gauss Lemma : "Let $a$, $b$ and $c$ three integers. If $\gcd(a,b)=1$ and if $a$ divides $bc$ then $a$ divides $c$."

I am working on generalizations of this lemma.

The first one is : "Let $a$, $b_1$,...,$b_n$ integers. If it exists $i \in \{1,...,n\}$ such as $\gcd(a,b_i)=1$ and if $a$ divides $\prod\limits_{k=1}^{n} b_k$ then $a$ divides $\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k$."

For the proof I use Bézout theorem to obtain : $au+b_iv=1$ then I multiply the equation by $\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k$. I obtain : $au\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k +v\prod\limits_{k=1}^{n} b_k=\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k$. Using the hypothsesis, I have : $a(u\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k +vQ)= \prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k$ where $Q \in \mathbb{Z}$. That is why $a$ divides $\prod \limits_{\substack{k=1 \\ k \neq i}}^{n} b_k$.

The second one is : "Let $a$, $b_1$,...,$b_n$ integers. If forall $i \in \{1,...,n-1\}$, $\gcd(a,b_i)=1$ and if $a$ divides $\prod\limits_{k=1}^{n} b_k$ then $a$ divides $b_n$."

For the proof, I use Bézout again to obtain $(n-1)$ equations of the form $au_i+b_i v_i=1, \ \forall i \in \{1,...,n-1\}$.

I multiply all these equations to have $\prod \limits_{i=1}^{n-1} (au_i+b_i v_i)=1$. Then I notice by developing the expression that I have $aK+b_1...b_{n-1}\prod \limits_{i=1}^{n-1} v_i=1$ where $K \in \mathbb{Z}$. I notice that if I have forall $i \in \{1,...,n-1\}$, $\gcd(a,b_i)=1$, I have $\gcd(a,\prod \limits_{i=1}^{n-1}b_i)=1$. To conclude I use the initial Gauss Lemma (with three integers).

For the last one I was wondering if it can work : "Let $a$, $b_1$,...,$b_n$ integers. If $\gcd(a,b_1,...,b_{n-1})=1$ and if $a$ divides $\prod\limits_{k=1}^{n} b_k$ then $a$ divides $b_n$."

Indeed, I tried to use Bézout generalized to have $au_0+b_1 u_1+...+b_{n-1} u_{n-1}=1$ but it does not seem to work like the previous lemma. So with this hyspothesis does this generalization can hold ?

Thanks in advance !

Maman
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  • The first is overdone, it is a direct conseqience of what you called Gauss's Lemma. Let $c$ be the product of all the $b_k$ except $b_i$. – André Nicolas Nov 29 '15 at 01:30
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    For your last lemma : if you take $n=3$, and $a=2$, $b_1=2$, $b_2=3$ and $b_3=1$, then you have $\gcd(a,b_1,b_2) = 1$, $a$ divides $b_1 b_2 b_3$ but $a$ does not divide $b_3$ – Esperluet Nov 29 '15 at 01:33
  • @AndréNicolas of course but I wanted to write it in a formal way ! – Maman Nov 29 '15 at 01:41
  • @Esperluet Indeed nice counter-example, this generalization cannot work ! – Maman Nov 29 '15 at 01:42
  • For some generalizations see the links in the last paragraph of this answer (Four-number Theorem, Schreier refinement and Riesz interpolation). The above is normally called Euclid's Lemma, not Gauss's. – Bill Dubuque Dec 01 '15 at 17:10
  • @BillDubuque There are countries which distinguish Euclid and Gauss lemma – Maman Dec 01 '15 at 21:46
  • @Maman Euclid already had the Four Numbers Theorem, and what you call Gauss's Lemma is a special case of that. – Bill Dubuque Dec 01 '15 at 22:32

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