Is x^y the Greatest Common Divisor of (5x+3y) and (13x+8y)?

mtayab1994
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Homework Statement



prove that: x^y=(5x+3y)^(13x+8y)

Homework Equations


The Attempt at a Solution



Can I say that x^y divides both 5x+3y and 13x+8y and go on from there or what?

Then in case one u could multiply 5x+3y by 13 and 13x+8y by 5 and do the difference and you'll get that x^y divides y

Case 2: multiply 5x+3y by 8 and 13x+8y by 3 and then we get x^y divides x.

And from case 1 and case 2 we can conclude that x^y=(5x+3y)^(13x+8y).

Note that ^ stands for the greatest common divisor.
 
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I strongly recommend that you NOT try to prove things that are not true!

Now, what is the problem really? For one thing, 8^{21} is not equal to 1.

Are you trying to prove that x^y is the greatest common divisor of (5x+ 3y)^{13x+ 8y}? Unfortunately, that's still not true. 13^{34}]/itex] is not divisible by 2.
 
HallsofIvy said:
I strongly recommend that you NOT try to prove things that are not true!

So that is not true or what?
 
Is this the question?:
Prove that the greatest common divisor of 5x+3y and 13x+8y is the same as the greatest common divisor of x and y.

or in notation I would understand:

Prove that gcd(5x+3y,13x+8y) = gcd(x,y)

And I suggest applying Euclid's algorithm to the polynomials on the left.
 
Joffan said:
Is this the question?:
Prove that the greatest common divisor of 5x+3y and 13x+8y is the same as the greatest common divisor of x and y.

or in notation I would understand:

Prove that gcd(5x+3y,13x+8y) = gcd(x,y)

And I suggest applying Euclid's algorithm to the polynomials on the left.

Yes I've solved it already thank you.
 
Good... I hope your solution looked something like:


Since ##\text{gcd}(m,n) = \text{gcd}(m-n,n)##,
<br /> \begin{align}<br /> \text{gcd}(13x+8y,5x+3y) &amp;= \text{gcd}(8x+5y,5x+3y)\\<br /> &amp;= \text{gcd}(3x+2y,5x+3y)\\<br /> &amp;= \text{gcd}(5x+3y,3x+2y)\\<br /> &amp;= \text{gcd}(2x+y,3x+2y)\\<br /> &amp; \dots<br /> \end{align}<br />etc.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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