Prove this relation between two numbers (Number theory)

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Homework Help Overview

The discussion revolves around proving a relation in number theory involving two integers, \( u \) and \( v \), specifically that if \( (u,v) = 1 \), then \( (u+v, u-v) \) is either 1 or 2. Participants are exploring the implications of coprimality and the definitions involved in the problem.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to understand the definitions of coprimality and the notation used. There are questions about the meaning of \( (u,v) = 1 \) and how it relates to Bézout's identity. Some are exploring the implications of specific integer examples, while others are questioning the necessity of defining \( u \) and \( v \) more clearly.

Discussion Status

The discussion is ongoing, with participants actively questioning assumptions and definitions. There is a mix of interpretations regarding the relationship between the integers and the notation used. Some participants have offered insights into the implications of the definitions, while others are still seeking clarity on the foundational concepts.

Contextual Notes

There is a noted lack of clarity regarding the definitions of \( u \) and \( v \), as well as the notation \( (u,v) \). Participants are also discussing the implications of the integers being coprime and how this affects the proof of the relation in question.

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Homework Statement
If (u,v) = 1, prove that (u+v,u-v) is either 1 or 2.
u and v are integers.
Relevant Equations
.
If (u,v) = 1, prove that (u+v,u-v) is either 1 or 2.
Where (,) means:
1609068130711.png


$$ux_1 + vx_2 = 1$$
$$u + v(x_2/x_1) = 1/x_1, u(x_1/x_2) + v = 1/x_2$$
$$u + v = 1/x_1 + 1/x_2 - v x_2/x_1 - u x_1/x_2$$
$$u - v = 1/x_1 - 1/x_2 + u x_1/x_2 - v x_2/x_1$$

Now we can express (u+v,u-v). But i am not sure if it will give us an answer.
I think the key is to know what x1 and x2 are, knowing that u and v are coprimes. But i don't know how to find it.
 
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I am missing the
Homework Statement:
and
Relevant Equations:

For part of the first I can substitute " If (u,v) = 1, prove that (u+v,u-v) is either 1 or 2."
But one really needs a definition of (u,v).
And one needs to know what u and v are. Uruguaians, Venezolans ?
 
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BvU said:
I am missing the
Homework Statement:
and
Relevant Equations:

For part of the first I can substitute " If (u,v) = 1, prove that (u+v,u-v) is either 1 or 2."
But one really needs a definition of (u,v).
And one needs to know what u and v are. Uruguaians, Venezolans ?
u and v are integer numbers hahaha
You are right, i forgot to write the meaning of (,).
I will edit
 
I don't understand how that definition of (u,v) that is given in post #1 is equivalent to that u,v are co primes.
 
Delta2 said:
I don't understand how that definition of (u,v) that is given in post #1 is equivalent to that u,v are co primes.
(u,v) = 1
ux + vy = 1 (1)​

" Bézout's identity — Let a and b be integers with greatest common divisor d. Then there exist integers x and y such that ax + by = d. More generally, the integers of the form ax + by are exactly the multiples of d. "

if a and b are co primes, gcd = 1
ax + by = 1

As in our case (1)
 
Sorry there is something i don't understand, let's take 5 and 2 that are coprimes, ##5\times10+2\times25=100\neq 1##

Something that you don't explain well, you saying for every ##x_1,x_2## or do you mean there exist ##x_1,x_2## such that
 
1609075102311.png

u and v are integers. Interesting.

Then what does it mean if you write "(u,v)=1" ?
That the set of all integers of the form ux1 + vx2 consists of one and only one element, namely 1 ?
 
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1 here represents the ideal generated by 1, or to put it a bit simply the set of all multiples of 1. This is equivalent to saying 1 is the smallest natural number that can be represented this was.

I think the solution to this is once you have a way of showing (u,v)=1, it's easy to give a representation that shows (u+v,u-v) contains 2, and hence is either 2 or 1.
 
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