Bézout's identity proof that if $(a,b,c)=1$ then $ax+bxy+cz=1$ has integer solutions

Question

Massive edit to simplify the question. Some comments below might be made obsolete - specifically, the comment that this follows directly from Dirichlet. That was true for the original wording.

I'm looking for a short proof, directly from Bézout's identity, of the following theorem:

Theorem 1: If $(a,b,c)=1$ then there exists an integer solution $x,y,z$ to $ax+bxy+cz=1$.

The case $(a,b)=1$ turns out to be equivalent to the theorem:

Theorem 2: The natural map: $$\mathbb Z_{nm}^\times\to\mathbb Z_{n}^\times$$ is onto.

That's because "onto" means if $(a,n)=1$ then for some $y$, $a+ny\in\mathbb Z_{mn}^\times$, meaning that $(a+ny,m)=1$ and thus $1=(a+ny)x+mz=ax+nxy+mz$ has a solution. The converse is equally obvious.

The general case in the first theorem follows if we know the case when $(a,b)=1$ since, for general $(a,b)$, we have $\left(\frac{a}{(a,b)},\frac{b}{(a,b)}\right)=1$, so from the special case, we get: $$\frac{a}{(a,b)}x_0 + \frac{b}{(a,b)} x_0y_0 + cz_0= 1$$ which implies:

$$ax_0 + bx_0y_0 + c((a,b)z_0)=(a,b)$$ Since $1=(a,b,c)=((a,b),c)$ we can find $(u,v)$ so that: $$(a,b)u + cv = 1$$

We then get:

$$a(x_0u) + b(x_0u)y_0 + c((a,b)z_0u + v) = 1$$

So $(x,y,z)=(x_0u,y_0,(a,b)z_0u+v)$ is a solution for our original equation. (Thanks Patrick Da Silva for that reduction.)

I can easily prove Theorem 2 using the structure of $\mathbb Z_n^\times$ in terms of prime factorizations. Indeed, Theorem 2 was the motivation for this question, initially - at first I thought it was "obvious," but the immediately realized it wasn't absolutely trivial.

It's certainly possible to translate the "abstract" proof of Theorem 2 into a direct proof of the special case of Theorem 1 using prime factorizations and Chinese remainder theorem.

But something about this theorem rang a bell for me. It looks like the sort of theorem that would have a short Bezout's identity proof.

Both unique factorization and Chinese remainder theorem are actually direct results of Bézout, and often theorems that we intuitively understand in terms of unique factorization and/or Chinese remainder theorem have a short, sharp proof using Bézout that eschews both the words "prime" and "remainder."

My instinct is that there ought to be a quick proof of the above with Bézout, without calling out to primes or remainders, but I haven't found it.

It's trivial if $(a,c)=1$, since $ax+cz=1$ lets us use $y=0$ to get a solution to $ax+bxy +cz=1$.

It's a little harder to see if $(b,c)=1$, but still not hard, since if $bu+cv=1$ then $$a\cdot 1 + 1\cdot (1-a) = a\cdot 1 + b(u(1-a)) + c(v(1-a))$$ giving a solution $(x,y,z)=(1,u(1-a),v(1-a))$.

That asymmetry (it's easy to solve if $(a,n)=1$ and harder to solve if $(b,n)=1$) suggests I might be wrong about there being such a proof, since Bézout is such a symmetric statement.

If there was a proof, it seems like you ought to start with:

$$au+bv=1\\ax+ny=(a,n)\\bw+nz = (b,n)$$

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As an example of a theorem that is "obvious" with unique factorization, but has a simple proof with Bézout's identity, consider:

$(a,n)=(a,m)=1\implies (a,mn)=1$

That has a unique factorization proof, but it follows directly from Bézout by multiplying: $$1=(ax_1+ny_1)(ax_2+my_2) = a(ax_1x_2 + mx_1y_2+nx_2y_1) + mn(y_1y_2)$$

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So, again, the goal is to have nothing about primes or Chinese Remainder Theorem in the proof, and to have it be "remarkably brief" - as much as possible, it shouldn't be hiding proofs of CRT or unique factorization.

I don't know that such a proof exists, but some instinct told me it did.

Answer 1

This may not be as simple as hoped, but it is pure Bezout.

We will use a couple of the results proven in this answer: $$ (ac,bc)=c(a,b)\tag{1} $$ and $$ (a,b)=1\quad\text{and}\quad(a,c)=1\implies(a,bc)=1\tag{2} $$ Furthermore, since $(a,b)\mid a$ and $(a,b)\mid bc$, we have $$ (a,b)\mid(a,bc)\tag{3} $$

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Lemma 1: If $(a,b)=1$, then $$ \begin{align} 1.&(a,n)(b,n)\mid n\\ 2.&(a+b,a^mb^n)=1 \end{align} $$ Proof: Suppose that $(a,b)=1$. Then for some $x,y$ we have $$ ax+by=1\tag{4} $$ Thus, we have $$ \begin{align} n &=n(ax+by)\\ &=\left(\frac{n}{(b,n)}\frac{ax}{(a,n)}+\frac{n}{(a,n)}\frac{by}{(b,n)}\right)(a,n)(b,n)\tag{5} \end{align} $$ and therefore, we conclude that $$ (a,n)(b,n)\mid n\tag{6} $$ Furthermore, from $(4)$, we also have $$ \begin{align} 1&=a(x-y)+(a+b)y&&(4)-ay+ay\tag{7}\\ 1&=(a+b)x+b(y-x)&&(4)+bx-bx\tag{8}\\ \end{align} $$ $(7)$ and $(8)$ say that $(a+b,a)=1$ and $(a+b,b)=1$. Repeatedly applying $(2)$ yields $$ (a+b,a^mb^n)=1\tag{9} $$ $\square$

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Lemma 2: $$ \left(a,\frac{n}{(a^n,n)}\right)=1 $$ Proof: Since $(a^k,n)\mid(a^{k+1},n)\mid n$, we must have, for some $k\le n$, that $(a^{k+1},n)=(a^k,n)$.

Suppose that $(a^{k+1},n)=(a^k,n)$. Then $\left(a\frac{a^k}{(a^k,n)},\frac{n}{(a^k,n)}\right)=1$ and therefore, $\left(a,\frac{n}{(a^k,n)}\right)=1$.

Since $k\le n$, we have $(a^k,n)\mid(a^n,n)$, and consequently, $$ \left(a,\frac{n}{(a^n,n)}\right)=1\tag{10} $$ $\square$

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Theorem: If $(a,b)=1$, then $$ \left(a+\frac{nu}{(a^n,n)(b,n)}b,n\right)=1 $$ where $u$ satisfies $(b,n)=bu+nv$.

Proof: Suppose that $(a,b)=1$ and $(b,n)=bu+nv$. Using $(2)$ repeatedly, we get $$ (a^n,b)=1\tag{11} $$ Combining $(6)$ and $(11)$ yield $$ (a^n,n)(b,n)\mid n\tag{12} $$ Thus, $\dfrac{n}{(a^n,n)(b,n)}\in\mathbb{Z}$ and $(10)$ says that $$ \left(a,\frac{n}{(a^n,n)(b,n)}(b,n)\right)=1\tag{13} $$ $(9)$ and $(13)$ imply that $$ \left(a+\frac{n}{(a^n,n)(b,n)}(b,n),a^n\frac{n}{(a^n,n)(b,n)}(b,n)\right)=1\tag{14} $$ Since $(a^n,n)\mid a^n$, we have $\left.n\,\middle|\,a^n\dfrac{n}{(a^n,n)(b,n)}(b,n)\right.$, so using $(3)$ and $(14)$ yields $$ \left(a+\frac{n}{(a^n,n)(b,n)}(b,n),n\right)=1\tag{15} $$ The rest is Bezout. $(15)$ says that there are $x,y$ so that $$ \left(a+\frac{n}{(a^n,n)(b,n)}(b,n)\right)x+ny=1\tag{16} $$ Furthermore, since $(b,n)=bu+nv$, we get $$ \left(a+\frac{nu}{(a^n,n)(b,n)}b\right)x+n\left(y+\frac{nv}{(a^n,n)(b,n)}\right)=1\tag{17} $$ which just says that $$ \left(a+\frac{nu}{(a^n,n)(b,n)}b,n\right)=1\tag{18} $$ $\square$

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A note on application

At first, $(18)$ may seem like a theoretical answer in the sense that it shows that we can find a $k$ so that $(a+bk,n)=1$, but computationally, it might appear a bit daunting due to the appearance of $(a^n,n)$ in the denominator. However, $n/(a^n,n)$ is simply $n$ with all the factors common to $a$ removed. To compute $n/(a^n,n)$, we can start with $n_0=n$, and generate the sequence $$ n_{k+1}=\frac{n_k}{(a,n_k)}\tag{19} $$ at some point $(a,n_k)=1$. For that $k$, we have $$ \frac{n}{(a^n,n)}=n_k\tag{20} $$

Answer 2

We use a constructive form of Stieltjes method of constructing coprimes.

${\bf Theorem}\quad\ \ \ \color{#c0f}{(a,b,c)=1}\, \Rightarrow\, (a\!+\!bn,c) = 1\,\ $ for some $\,n\,$ that satisfies $\ \color{#c00}{(n,a)=1}$

$\require{cancel}{\bf Proof}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \phantom{1^{1^{1^{1^{1^{1^1}}}}}} \phantom{\dfrac{1}{1^{1^1}}} (a,c) = 1\,\Rightarrow\, \smash[t]{\overbrace{(a\!+\!bc,c)}^{\large (a,\,c)}=1.}\ $ Else induct on $c\!:\ $ $(a\!+\!bn,\color{#c70}{\large \frac{c}{(a,c)}})=1,\,$ $\, \color{#c00}{(n,a)\!=\!1},\,$ $(a\!+\!bn,\color{#0a0}{(a,c)}) = (bn,(a,c))=1\,$ by $\,\color{#c0f}{(b,a,c)}=1=(\color{#c00}{n,a},c)\ $ so $\ (a\!+\!bn,\color{#C70}{\large \frac{c}{\cancel{(a,c)}}}\!\color{#0a0}{\cancel{\small (a,c)}}\!)=1$

Answer 3

Application of Chinese Remainder Theorem:

$$x\equiv a \textrm{ (mod $b$)}$$ $$x\equiv 1\textrm{ (mod $p$)}, \textrm{ for primes $p$ that divides $n$, but does not divide $b$}$$

Answer 4

A correct but imperfect answer, loosely based on the ideas in robjohn's answer.

Lemma 1: If $(a,b)=1$ then for any $n$, we can find $n=n_1n_2$ with $(n_1,n_2)=(a,n_1)=(b,n_2)=1$.

Proof: This requires induction. If $(n,a)=1$ then $n_1=n$ and $n_2=1$. Otherwise, let $m=\frac{n}{(a,n)}$. Then by induction $m=m_1m_2$ with $(m_1,m_2)=(m_1,a)=(m_2,b)$. But then let $n_1=m_1$ and $n_2=m_2(a,n)$. Since $(a,b)=1$, $((a,n),b)=1$, so $(n_2,b)=1$. So we're done.

[That assumes $(x,z)=(y,z)=1\Rightarrow (xy,z)=1$, which is easy to prove with Bézout.]

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Theorem 1: If $(a,b)=1$ then $\exists x,y.z: ax+bxy + cz=1$

Proof: By Lemma 1, we can write $c=c_1c_2$ with $(c_1,c_2)=(a,c_1)=(b,c_2)=1$. Solve the following equations:

$$ c_1p + c_2q = 1\tag{1}$$ $$au_1+c_1v_1=1\tag{2}$$ $$bu_2+c_2v_2=1\tag{3}$$

Rewrite (3) as: $$a\cdot 1 + bu_2(1-a) + c_2v_2(1-a)=1\tag{4}$$ Multiplying (2) by $c_2q$ and (4) by $c_1p=1-c_2q$ and adding, we get:

$$a(c_2qu_1 + 1-c_2q) + b(u_2(1-a)c_1p) + c_1c_2(v_1q+v_2p(1-a))=1\tag{5}$$

Letting $y=u_2(1-a)c_1p, x=c_2q(u_1-1)+1, z_0=v_1q+v_2p(1-a)$ we see that:

$$ax + by+ cz_0=1\tag{6}$$

And here's the magic: $$xy = c_2q(u_1-1)y + y = c_1c_2q(u_1-1)p(1-a)u_2 + y=cD+y\tag{7}$$

where $D=q(u_1-1)p(1-a)u_2$.

So $$ax + bxy + c(z_0-bD)=1$$This proves our theorem.

Not entirely satisfying, except for the fact that we have a systematic way to find $x,y,z$ for any $(a,b,c)$.

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What is going on here? As noted in the question, if $(a,c)=1$ then $ax+cz=1$ has a solution, and $y=0$ gives us our solution.

If $(b,c)=1$, then solving $bu+cv=1$ which gives us a solution:

$$a\cdot 1 + b(u(1-a)) + c(v(1-a))=1$$ So above, we have essentially easy solutions to:

$$ax+bxy \equiv 1 \pmod {c_1,c_2}$$

and we are combining these easy solutions to these two equations in the way that Bézout's identity lets us prove Chinese Remainder Theorem.

The (not-really-so-magic) fact that $xy\equiv y\pmod {c}$ comes from the fact that $x\equiv 1\pmod {c_2}$ and $y\equiv 0\pmod {c_1}$.

Lemma 1 is essentially robjohns work about $(a^n,n)$, but in reverse. I think Lemma 1 is a more direct induction - exponentiation is artificial in this problem.

This proof is also pretty much exactly the same answer as i707107, who couched the answer in terms of prime factorization and Chinese Remainder Theorem.

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Related But Possibly Less Mysterious Proof

There is a slightly more direct approach where we can prove a more specific theorem:

Theorem 3: If $(a,b)=1$ then for any $c$ we can find $x,y,z$ so that:

$$ax+by+cz = 1\\ c\mid xy-y$$

Note: Theorem 3 clearly implies Theorem 1. It is not transparent that Theorem 1 implies Theorem 3.

Proof of Theorem 3:

As noted before, when $(a,c)=1$ we can find a solution to the linear equation with $y=0$. If $(b,c)=1$, we can find a solution with with $x=1$. In both cases, $xy-y=0$ so the second condition is true.

So all we need to show is that if $(c_1,c_2)=1$ and we have solutions:

$$ ax_i+by_i+c_iz_i=1\\c_i\mid x_iy_i-y_i$$

for $i=1,2$, then we can find a solution:

$$ax+by+c_1c_2z = 1\\ c_1c_2\mid xy-y$$

The direct way to combine the linear equations just works. Solve $c_1u+c_2v=1$ and choose:

$$x = c_2vx_1 + c_1ux_2\\y=c_2vy_1 + c_1uy_2\\z=z_1v+z_2u$$

To get $c_1c_2$ we multiply the first linear equation by $c_2v$ and the second by $c_1u$ and add to get a linear equation for $a,b,c_1c_2$.

Now you just need to verify that $xy-y$ is divisible by $c_1c_2$. This can be done by painful calculation. The big trick is using that $(c_1u)^2 = c_1u(1-c_2v) = c_1u - c_1c_2(uv)$ and likewise that $(c_2v)^2 = c_2v - c_1c_2(uv)$. This lets you get that $xy = c_2vx_1y_1 + c_1ux_2y_2 + c_1c_2D$ for some $D$. But then $x_iy_i-y_i$ divisible by $c_i$ means that we can write this as: $xy = c_2vy_1 + c_1uy_2 + c_1c_2D_2 = y + c_1c_2D_2$, for some $D_2$.

Essentially, the reason that $c_1c_2\mid xy-y$ is that $x$ and $y$ are the formulas for the Chinese remainder theorem, so $x\equiv x_i\pmod {c_i}$ and hence $c_i\mid xy-y$ for each $i$. But you can avoid appeals to the remainder theorem by just working it out and proving that $c_1c_2|xy-y$.

Once you have this result, you can use Lemma 1 above to prove the general result for all $c$.

One last comment. Lemma 1 actually is easily generalized to:

Lemma 2: If $(a,b,c)=1$ then there exists a factoring $c=c_1c_2$ with $(c_1,a)=(c_2,b)=(c_1,c_2)=1$.

Proof: Letting $a'=\frac{a}{(a,b)}$ and $b'=\frac{b}{(a,b)}$ we apply Lemma 1 to $a',b',c$ we get $c=c_1c_2$ with $(c_1,c_2)=(a',c_1)=(b',c_2)=1$. But then since $((a,b),c_1)=1$ and $(a',c_1)=1$, we hve that $(a,c_1)=1$. Similarly for $b$. This requires the Lemmas:

Lemma 3: If $(a_1,c)=(a_2,c)=1$ then $(a_1a_2,c)=1$.

Proof: This is easily proved from Bézout by multiplying the two equations $a_ix_i+cy_i=1$ for $i=1,2$.

Lemma 4: If $(a,b)=1$ and $c\mid b$ then $(a,c)=1$.

Proof: Again, easy to prove from Bézout's - if $c=bd$ then $1=ax+by=ax+c(dy)$.

Lemma 2 means that we can remove the condition that $(a,b)=1$ in our theorems - we only used that condition to factor $c$ via Lemma 1.

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Robjohn's proof, simplified

Robjohn's answer can be summarized as follows:

Lemma 5: If $(a,b,n)=1$ then there exists an $n_1$ such that the following conditions hold: $$(a+n_1,n)=1\\(b,n)\mid n_1\mid n$$

Proof: Induction on $n$.

If $(a,n)=1$ we can choose $n_1=n$ since $(a,n)=1\Rightarrow (a+n,n)=1$ and $(b,n)\mid n$.

If $(a,n)>1$ write $n=(a,n)m$, and find $m_1$ so that $(a+m_1,m)=1$ and $(b,m)\mid m_1\mid m$. . Writing $$1=(a+m_1)x + my = ax + m_1\left(x+\frac{m}{m_1}y\right)$$, we see that $(a,m_1)=1$.

Let $n_1=m_1$. Then, $(a,n_1)=(a,m_1)=1$, and we can conclude that $(a+n_1,a)=1$ so $(a+n_1,(a,n))=1$. So $$(a+n_1,n)\mid (a+n_1,(a,n))\cdot(a+n_1,m)=1$$

Finally, we can easily see that, since $(a,n)=1$, that $(b,n)=(b,m)$. So $(b,n)=(b,m)\mid n_1=m_1\mid m\mid n$.

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Theorem: If $(a,b,n)=1$ then there exists an $x,y,z$ so that $ax+bxy+cz=1$.

Proof: Let $n_1$ be as in Lemma 5. Then we can find solutions to: $$bu+nv=n_1\\(a+n_1)x+nz=1$$

Replacing $n_1$ in the second, we get: $$(a+bu)x + n(z+xv)=1$$

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This is the heart of robjohn's solution, but it avoids exponentiation in the argument, and minus all the fractions. Essentially, $n_1=\frac{n}{(a^n,n)}$, but those details are noisy for the rest of the proof, while these are the only properties we really need for $n_1$ and we can prove that by induction directly.

Answer 5

I think I've found an excellent short proof.

Theorem 1: If $(a,b,c)=1$ then there exists $x,y,z$ so that $$ax+bxy+cz=1$$.

Proof: Induction on $c$ and $b$.

We can solve if $b=1$ by setting $(x,y,z)=(1,(1-a),0)$.

Now, assume that, for some $c$ we can solve for $(a,b,c')$ if $c'<c$ and $(a,b',c)$ if $b'<b$.

Then we break this up into two cases.

Case 1 If $b\not\mid c$, then $(b,c)<b$ and, by the induction hypothesis, we can solve:

$$ax+(b,c)xy+cz =1$$ and $$bu+cv=(b,c)$$

Then $$ax +bx(uy) + c(z+vxy)=1$$

Case 2 If $1<b\mid c$, then $1=(a,b,c)=(a,b)$. Let $c=bd$.

Since $d<c$ and $(a,b,d)|(a,b,c)=1$, we can solve:

$$ax+bxy + dz=1$$

That can be seen as saying that $(a+by,d)=1$. But from $(a,b)=1$, we can see that $(a+by,b)=1$. So this means that $(a+by,c)=(a+by,db)=1$. And then we can find a solution to $$(a+by)X+cZ=1$$

(We could be more explicit in that last step: If $au+bv=1$ then $(a+by)u + b(v-yu)=1$. Multiplying that by $(a+by)x+ dz=1$ gives us an explicit solution to $(a+by)X + bdZ=1$.)

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The trick here in case (1) was in robjohn's proof - he solved for the triple $(a,(b,c),c)$ first, then used that to show there was a solution for the triple $(a,b,c)$. I've just pushed that trick into the descent part of the proof.

This hides the "factoring" part of the proof. We are certainly essentially factoring out the parts of $c$ that are common with $b$ in the descent, but it feels more organic.

An interesting fact about this proof is that the new $y$ we get via induction is always the same $y$ or $yu$. Since $a$ doesn't change in the induction, and $y=1-a$ when $b=1$, this means that the $y$ we get from this proof is always divisible by $a-1$.