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I have tried to think as intersections of two lines but for $2x+3$ and $5x+7$ is failed to have integer intersections (but not sure it is enough counterexample).

Is the above generalisation correct? Can you give a counter-example, or a hint for how to prove it?

Jale'de jaled
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    you have $ 2 * 2 + 3 = 5 * 0 + 7$. The "x" need not be the same. – Calvin Lin Jan 03 '21 at 16:39
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    You assumed slightly too much. It's not that $2x+3 = 5x+7$ but rather that $2x+3 = 5y+7$ for some natural numbers $x,y$. – Cameron Williams Jan 03 '21 at 16:40
  • Immediate consequence of CRT, e.g. apply the Theorem in the dupe to $,z\equiv b\pmod{a},\ z\equiv y\pmod{x}\ \ $ – Bill Dubuque Jan 03 '21 at 17:14
  • "Immediate consequence of CRT" minor additional step. CRT says $w\equiv b\pmod a; w\equiv y\pmod x$ has unique solutions $\pmod {ax}$ so there exist integers $M,N$ so that $w = aM + b=xN + y$. But it doesn't say $M,N$ are natural numbers. But that is an exceedingly minor point. ..... $w+k(ax) = a(M+kx) =x(N+ka)$ so we can add multiples of $x$ and $a$ to $M,N$ until they are both positive. – fleablood Jan 03 '21 at 17:42

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This is true, and in fact you only need the first condition $\gcd(a,x)=1$. This follows from the Chinese remainder theorem. For example, $2\mathbb N+3$ and $5\mathbb N+7$ both include all points in $10\mathbb N+7$.

Especially Lime
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Hint: Find integers $A, X$ such that $ a A +xX = y - b$.

Calvin Lin
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  • To show $A$, $X$ exist: $gcd(x,a) = 1$, then $\alpha x+\beta a = 1$, for some $\alpha, \beta$ .Then multiply both sides by $(y-b)$. – Aven Desta Jan 03 '21 at 16:47
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First since $gcd(x,a)=1$, there exist $\alpha$, $\beta$, such that (Bézout's identity)$$\alpha x - \beta a = 1$$ Now multiply both sides by $k = b-y$.

$$(k \alpha ) x - (k \beta)a = b-y$$ $$(k \alpha ) x + y = (k \beta)a +b$$

Aven Desta
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Thinking in terms of straight lines and looking for points of intersection where both lines have the same output for the same input is not the best analogy. It's better to think of these as unstructured unordered sets which are like bags of groceries that they have a bunch of stuff dumped in them in no particular order and an "intersection" simply means the two bags both have the same item somewhere. ("You bought mushrooms and tampons too! So did I!")

$2\mathbb N + 3 = \{5,7,9,11,13,15,17,19,21,23,25,27,29......\}$ and $5\mathbb N + 7= \{12,17,22,27,32,....\}$

Notice they both have $17$ and $27$ and $37$ etc in them. Note those numbers aren't in both for the same reason... If $17 \in 2\mathbb N+3 = \{2\color{blue}n + 3|\color{blue}nn\in \mathbb N\}$ it is because if $\color{blue}{n= 7}$ we get $2\cdot\color{blue}{7} +3=17$ which $17 \in 5\mathbb N + 7=\{5\color{red}n + 7|\color{red}n \in \mathbb N\}$ because if $\color{red}n = 2$ we get $5\cdot\color{red}2+7=17$... but we don't care why the items are in the bag, just that they are both in each of the bags.

("I bought the mushrooms as the fifth item right after the milk and a picked up the tampons at the check out counter as my last item" "Oh, not me. I went shopping just because I needed tampons so I bought the first thing. Then I thought about dinner so I bought pasta, mushrooms, and onions. Mushrooms were the third thing I bought.")

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So this is actually much looser.

You mus show that there exist natural number $M,K$ so that $aM + b= xK +y$ where $M,K$ can be any natural numbers and need not have any thing in common.

That would mean $aM - xK = (y-b)$. By Bezout's lemma you know we can find integers$M', K'$ so that $aM' + xK' = 1$ so if we let $M = (y-b)M'$ and $K = -(y-b)K'$ we get $aM - xK = y-b$.

Or only concern is whether or not $M,K$ are both positive. I'll let you work the details.

(Note: if $a,b \ge 0$ then $aM' + bK' = 1$ would mean only one of $M'$ or $K'$ is positive and the other is either negative or $0$.)

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Postsript: For the record the two sets $a\mathbb N + b$ and $x \mathbb N + y$ don't just intersect and have one item in common. They have an infinite number of items in common.

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Notice that every $10K + 7 = 2(5K + 2) + 3= 5(2K) + 7 \in (2\mathbb N + 3)\cap( 5\mathbb N + 7)$. So for all $K$ we have $10K + 7$ is in the intersection.

That may give you a hint to a more direct way (but not necessarily easier) way to solve it than my Bezout solution.

fleablood
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