Add Me!Close Menu Navigation

Odd Coprimes

Let a be an odd integer. Prove that a^{2^n} + 2^{2^n} and a^{2^m}+2^{2^m} are relatively prime for all distinct positive integers n and m.

Because m,n are distinct, then we can impose an arbitrary ordering on them such that m > n. Suppose that for any prime p|a^{2^n}+2^{2^n}, we find that by definition

    \begin{align*} p|a^{2^n}+2^{2^n} &= p|a^{2^n}- (-2^{2^n}) \\ &\implies a^{2^n} \equiv -2^{2^n} \mod p \\ \intertext{because $m>n$, then we can square both sides $m-n$ times to get} (a^{2^n})^{2^{m-n}} \equiv (-2^{2^n})^{2^{m-n}} \mod p &\implies a^{2^m} \equiv 2^{2^m} \mod p \\ &\implies p|(a^{2^m} - 2^{2^m}) \end{align*}

Because a^{2^m} + 2^{2^m} = a^{2^m} - 2^{2^m} + 2\times 2^{2^m} \equiv 0 \mod p only when p = 2. However, since a is odd, p cannot be 2, therefore, p \centernot | a^{2^m} + 2^{2^m}. Because every prime factor of a^{2^n} + 2^{2^n} cannot divide a^{2^m} + 2^{2^m}, then by definition, these two expressions are relatively prime for any distinct n,m. \blacksquare

Posted By Lee

Woosh

Recent Comments