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Attached below is a statement and a proof from the book "104 Number Theory Problems" by Dorin Andrica, Titu Andreescu, and Zuming Feng.

Here is the corollary mentioned in the proof.

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The point I'm having trouble understanding is how they declared there exist $(u,v) \geq0$ such that $\gcd(x,y)=ux-vy$. From my understanding of Bezout's identity, there exist countably infinite integer solutions to $ax+by=\gcd(x,y)$, but how exactly is a positive-negative pair guaranteed? For positive integers $(x,y)$, I understand why this has to true as their gcd has to be less than or equal to both of them, hence one of the coefficients needs to be negative. When $(x,y)$ are both negative, it's possible to just flip the signs of the solution pair of $a(-x)+b(-y)=\gcd(-x,-y)$ as $\gcd(x,y)=\gcd(-x,-y)$. The cases where either $x$ or $y$ is negative are the ones I can't seem to justify.

On another note, why was this declaration required? The proof remains valid using integers $(a,b)$ as in $ax+by$ instead of $ux-vy$.

Any help is appreciated.

Typo
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  • If $x<0$ then $,a^x \equiv b^x\Rightarrow a,b,$ invertible so inverting $\Rightarrow a^{-x}\equiv b^{-x},$ reduces to positive exponent case. But it's not clear from the proof that they intend it to handle exponents $< 0,,$ i.e. they may intend $,x,y\in\Bbb N,,$ which would explain why they want $,u,v\ge 0.\ $ See the linked dupes for this method and others. – Bill Dubuque Jun 19 '22 at 15:42
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    Indeed, they are using the quoted lemma to cancel powers of $a$ and $b$ (vs. using inverses), i.e. they cancel $,a^{vy} b^{vy},$ from $,a^{ux} b^{vy} \equiv a^{vy} b^{ux}$ to get $,a^{ux-vy}\equiv b^{ux-vy}.,$ So it is likely done that way to avoid requiring any knowledge of inverses (and negative powers) at that point. – Bill Dubuque Jun 19 '22 at 16:07
  • @BillDubuque Thanks for the info and the links. I'll go over your responses and the other threads and let you know if I have any additional questions. – Typo Jun 19 '22 at 16:12
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    Note also that by homogeneous reduction we can reduce to the coprime case $,(x,y) = 1,$ as in the dupes, i.e. let $,d = (x,y),\ x = d\bar x,\ y = d\bar y,\ $ so $,a^x = (a^d)^{\bar x} =: A^{\bar x},,$ etc, so we reduce to the problem $,(\bar x,\bar y)=1,\ A^{\bar x}\equiv B^{\bar x},\ A^{\bar y}\equiv B^{\bar y}\Rightarrow A\equiv B\ \ $ – Bill Dubuque Jun 19 '22 at 16:46
  • @BillDubuque So I've gone over everything and I understand how the proof works, the point about the cancelling property is especially clearer now. Thanks for that. I still have some questions if you have the time. In your 1st comment, what did you mean by "reduces to positive exponent case" in relation to my question? And can you point me towards the explanation of positive-negative pair of Bezout solutions existing for the case of either $x$ or $y$ is negative? (Your first comment probably answers this but I don't see the connection unfortunately...) – Typo Jun 20 '22 at 00:35
  • I hadn't perused the book when I wrote the first comment, so it wasn't yet clear what method they were using in Cor. 1.21 (it would have been best to quote that). re: Bezout, as I said, they likely meant to assume $x,y$ nonnegative, and you already know how to handle that. – Bill Dubuque Jun 20 '22 at 12:48
  • @BillDubuque You are right, I should've mentioned it. I'll edit it in. re: re: Bezout, I am still curious about the property being true or not, hence I ask if you can kindly point me towards a hint or proof/counterexample. – Typo Jun 20 '22 at 13:50
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    It's best to ask a new question divorced from this "typo" context. It is clear that they intend $x,y$ to be $\ge 0$ since they don't introduce modular inverses till after that point in the book (so $a^x\equiv b^x\pmod m$ is not be well-defined if $x<0$ at that point in the book). – Bill Dubuque Jun 20 '22 at 14:03
  • Very well, I shall do that. – Typo Jun 20 '22 at 14:04
  • For the record, the "divorced" Bezout question was posted here (mentioned to get links in place). – Bill Dubuque Jun 21 '22 at 14:10

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