In mathematics, the Goormaghtigh conjecture is a conjecture in number theory named for the Belgian mathematician René Goormaghtigh about the solutions of the exponential Diophantine equation
x m − 1 x − 1 = y n − 1 y − 1 {\displaystyle {\frac {x^{m}-1}{x-1}}={\frac {y^{n}-1}{y-1}}}
with distinct integers x , y {\displaystyle x,y} larger than one and exponents larger than two. One convention is x > y > 1 {\displaystyle x>y>1} and in turn n > m > 2 {\displaystyle n>m>2} . The conjecture states that the only such solutions are
5 3 − 1 5 − 1 = 2 5 − 1 2 − 1 = 31 {\displaystyle {\frac {5^{3}-1}{5-1}}={\frac {2^{5}-1}{2-1}}=31}
and
90 3 − 1 90 − 1 = 2 13 − 1 2 − 1 = 8191. {\displaystyle {\frac {90^{3}-1}{90-1}}={\frac {2^{13}-1}{2-1}}=8191.}
Representation The fraction of either side of the conjecture exactly represents a finite geometric series. Indeed, x m − 1 x − 1 = ∑ k = 1 m x k − 1 {\displaystyle \textstyle {\frac {x^{m}-1}{x-1}}=\sum _{k=1}^{m}x^{k-1}} and so, for example, 31 = 1 + 5 + 25 = 5 0 + 5 1 + 5 2 {\displaystyle 31=1+5+25=5^{0}+5^{1}+5^{2}} . As such, the exponential Diophantine equation equates two univariate polynomials, with m {\displaystyle m} terms and highest order x m − 1 {\displaystyle x^{m-1}} on the left hand side, and n > m {\displaystyle n>m} on the right. Alternatively, by cross-multiplication of the fraction's denominators, the equation is equivalently expressed as
x m + x ⋅ y n + y = y n + y ⋅ x m + x , {\displaystyle x^{m}+x\cdot y^{n}+y=y^{n}+y\cdot x^{m}+x,}
or similar forms. Taking logs,
m n = ln y ln x + O ( 1 n ) {\displaystyle {\frac {m}{n}}={\frac {\ln y}{\ln x}}+O({\tfrac {1}{n}})}
where the remainder term is a log x {\displaystyle \log _{x}} of a ratio of polynomial expressions. Given x , y , n {\displaystyle x,y,n} , one has m = n ⋅ log x y + O ( 1 ) {\displaystyle m=n\cdot \log _{x}y+O(1)} with the remainder in the range ( − 1 , 1 ) {\displaystyle (-1,1)} .
… excerpt ends here. Continue reading the full article.
