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Request the needed missing step proof below.

Given $a, b\in \mathbb{N}, p\mid ab$, given a prime number $p$, then at least one of $a, b$ is divisible by $p$.

Proof: By hypothesis, $ab = cp$. Wlog, $p\not \mid a$. Then $(a, p)=1$. Hence, by Bezout lemma, there exist $x, y$, which satisfy $ax + py = 1$. Multiplying this equation by $b$, have $bax + bpy = b$.

Since $ab = cp$, this gives $cpx + bpy = b$, or $(cx + by)p = b$. Since $cx + by$ is an integer, but cannot still show the needed step that $(cx+by) \not \mid b$ to show $p\mid b$.

jiten
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    You don't need to, you showed that $b$ is equal to $p$ multiplied by some other integer, that's exactly what $p|b$ means – Alessandro May 14 '22 at 06:28
  • @Alessandro still something more is needed, as say $cx+by= mp'$, where $p'$ is a prime and $p'=p$. – jiten May 14 '22 at 06:33
  • No, Alessandro is exactly right. "$p$ divides $b$" means $b=pQ$ for some $Q$; you've shown this is true for $Q=cx+by$. – Gerry Myerson May 14 '22 at 06:37
  • @GerryMyerson Please take an example, say: $p=5, ab = 40, a= 8, b=5$, then $c=a=8, x=0$, $y= ?$, so that get $cx+by=0+1= 1$; leading to $1.p= p$. Trying, say for $ab= 200, p=5, a= 8, b=25$ fail still to find suitable $y$. – jiten May 14 '22 at 06:43
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    $8x+5y=1$ has the solution $x=2$, $y=-3$ (and infinitely many other solutions, but we only need one). What's wrong with $1\times p=p$? $8x+25y=1$ has the solution $a=-3$, $y=1$. – Gerry Myerson May 14 '22 at 07:06
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    I think the confusion is pretty basic. If $(cx + by)p = b$, both $cx + by$ and $p$ divide $b$. – sku May 14 '22 at 07:15
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    By definition $,p\mid b\iff b = \color{#c00}np,$ for some $,n\in \color{#0a0}{\Bbb Z},,$ which is true in your proof since $, b = (\color{#c00}{cx+by})p,$ so you can choose $,\color{#c00}{n =cx+by},$ since it is an $\rm\color{#0a0}{integer}$. See this Remark linked dupe for more on this form of Euclid's Lemma and related results. – Bill Dubuque May 14 '22 at 07:18
  • @GerryMyerson Thanks for showing, as that is what requested earlier, i.e. how to show for $ab=40, p=5, a=8, b=5$ have $x,y$ such that $cx+by= 1$. Have $c= \frac{ab} p = 8$, so need show $8x+5y=1$. You gave value $x=2, y=-3$. Get: $8.2-5.3= 1$. Other infinitely many solutions should be : $x'= x-5i, y'= y+8i$, or $x'= x+5i, y'= y-8i$ for integer $i$. Taking first approach, get for $i=1, x'= 2-5=-3, y'=-3+8=5$, which gets: $-24+25=1$. Taking second approach, get for $i=-1, x'= 2+5=7, y'=-3-8=-11$, which gets: $56-55=1$. – jiten May 14 '22 at 11:17

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