Difference between revisions of "Zsigmondy's Theorem"

Revision as of 03:07, 30 August 2025

Zsigmondy's Theorem states that, for positive relatively prime integers $a$ , $b$ , and $n$ with $a>b$ , there exists a prime number $p$ (called a primitive prime factor) such that $p|(a^n-b^n)$ but $p\not|(a^k-b^k)$ for all positive integers $k<n$ EXCEPT (i) if $n=a-b=1$ , (ii) if $n=2$ and $a+b$ is a power of $2$ , or (iii) if $n=6$ , $a=2$ , and $b=1$ . This theorem can sometimes be used to prove that no more solutions exist to Diophantine equations.

Example

We desire to find all solutions $(x,y)$ to the Diophantine $3^x-1=5^y$ . We notice that the first integer $x$ for which $5|3^x-1$ is $x=4$ , which produces $3^4-1=80\neq 5^y$ . Now, by Zsigmondy's Theorem (the exceptions do not apply here), for any positive integers $n\geq 4$ , there must exist a primitive prime factor for each of the terms $3^n-1^n.$ However, the right hand side of the equation only contains prime factors of $5$ , and so there are $\boxed{\text{no solutions}}$ to this Diophantine equation.

Problems

Find all solutions $(x,y,z)$ to the Diophantine equation $7^x-1=2\cdot3^y\cdot5^z$ .
Prove that the equation $(x+y)^n=x^m+y^m$ has a unique solution in integers with $x>y>0$ and $m,n>1$ .

@@ Line 8: / Line 8: @@
 * Find all solutions <math>(x,y,z)</math> to the Diophantine equation <math>7^x-1=2\cdot3^y\cdot5^z</math>.
+* Prove that the equation <math>(x+y)^n=x^m+y^m</math> has a unique solution in integers with <math>x>y>0</math> and <math>m,n>1</math>.
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Difference between revisions of "Zsigmondy's Theorem"

Revision as of 03:07, 30 August 2025

Example

Problems

See Also