If 
x
,
y
>
1
 but 
gcd
(
x
,
y
)
=
1
,  which of the following is not  true?

		

if 
a
|
x
 and 
a
|
y
 for positive integer 
a
, then 
a
=
1

		

if 
a
 is an integer, then 
a
=
s
x
+
t
y
 for some integers 
s
 and 
t

		

1
=
s
x
+
t
y
, for some integers 
s
 and 
t

		

x
k
=
(
x
−
y
)
, for some integer 
k

Given \( x, y > 1 \) and \(\gcd(x, y) = 1\), let's analyze each of the statements to determine which is not true:

1. **if \(a | x\) and \(a | y\) for positive integer \(a\), then \(a = 1\)**

    Since \(\gcd(x, y) = 1\), the only positive integer that divides both \(x\) and \(y\) is 1. This statement is true.

2. **if \(a\) is an integer, then \(a = sx + ty\) for some integers \(s\) and \(t\)**

    This statement is asserting that any integer \(a\) can be written in the form \(sx + ty\) where \(s\) and \(t\) are integers. This is a very general claim and does not hold true in general. For instance, if \(a = 2\), \(x = 3\), and \(y = 5\), there are no integers \(s\) and \(t\) such that \(2 = 3s + 5t\). Therefore, this statement is not true.

3. **\(1 = sx + ty\), for some integers \(s\) and \(t\)**

    This statement is true by Bezout's Identity, which states that for any two integers \(x\) and \(y\), there exist integers \(s\) and \(t\) such that \(sx + ty = \gcd(x, y)\). Since \(\gcd(x, y) = 1\), there exist \(s\) and \(t\) such that \(sx + ty = 1\).

4. **\(x^k = (x - y)\), for some integer \(k\)**

    For this equation to be true, we need \(x^k = x - y\). This would imply that \(x^k - x + y = 0\). Since \(x > 1\) and \(y > 1\), there is no integer \(k\) that can satisfy this equation for all \(x\) and \(y\). Therefore, this statement is not true.

Thus, the statements which are not true are:

- if \(a\) is an integer, then \(a = sx + ty\) for some integers \(s\) and \(t\)
- \(x^k = (x - y)\), for some integer \(k\)

The first incorrect statement encountered was the second one.

### Relative Questions:
1. Can you explain Bezout's Identity in more detail?
2. What are the implications of \(\gcd(x, y) = 1\) in number theory?
3. How do we know that \(x^k = (x - y)\) has no integer solution for \(k\)?
4. Can you provide an example to show why \(a = sx + ty\) does not hold for all integers \(a\)?
5. What other properties arise when two numbers are coprime (\(\gcd(x, y) = 1\))?

### Tip:
When working with coprime numbers, the greatest common divisor (gcd) being 1 provides a powerful tool for proving many useful properties, including those related to linear combinations and divisibility.

If x, y > 1 but gcd(x, y) = 1, which of the following is not true?

if a | x and a | y for positive integer a, then a = 1

if a is an integer, then a = sx + ty for some integers s and t

1 = sx + ty, for some integers s and t

x^k = (x − y), for some integer k

Explore which statements are not true when x, y > 1 and gcd(x, y) = 1. Dive into the properties of coprime numbers, Bezout's Identity, and divisibility rules. Suitable for college-level number theory students.

Math Problem Statement

1 but gcd ( x , y )

if a | x and a | y for positive integer a , then a

if a is an integer, then a

1

x k

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