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Binomial Number

A number of the form $a^n\pm b^n$, where $a,b$, and $n$ are Integers. They can be factored algebraically

\begin{displaymath}
a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})
\end{displaymath} (1)


\begin{displaymath}
a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\ldots-ab^{n-2}+b^{n-1})
\end{displaymath} (2)


\begin{displaymath}
a^{nm}-b^{nm}=(a^m-b^m)[a^{m(n-1)}+a^{m(n-2)}b^m+\ldots+b^{m(n-1)}].
\end{displaymath} (3)

In 1770, Euler proved that if $(a,b)=1$, then every Factor of
\begin{displaymath}
a^{2^n}+b^{2^n}
\end{displaymath} (4)

is either 2 or of the form $2^{n+1}K+1$. If $p$ and $q$ are Primes, then
\begin{displaymath}
{(a^{pq}-1)(a-1)\over (a^p-1)(a^q-1)}-1
\end{displaymath} (5)

is Divisible by every Prime Factor of $a^{p-1}$ not dividing $a^q-1$.

See also Cunningham Number, Fermat Number, Mersenne Number, Riesel Number, Sierpinski Number of the Second Kind


References

Guy, R. K. ``When Does $2^a-2^b$ Divide $n^a-n^b$.'' §B47 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, p. 102, 1994.

Qi, S and Ming-Zhi, Z. ``Pairs where $2^a-2^b$ Divides $n^a-n^b$ for All $n$.'' Proc. Amer. Math. Soc. 93, 218-220, 1985.

Schinzel, A. ``On Primitive Prime Factors of $a^n-b^n$.'' Proc. Cambridge Phil. Soc. 58, 555-562, 1962.




© 1996-9 Eric W. Weisstein
1999-05-26