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QUESTION

For each integer, $n\geq0$, define $h_n=2^{2^n}+1$.

\begin{description}

\item[(i)]
Evaluate $h_0,h)1,h_2,h_3$.

\item[(ii)]
Show that HCF($h_n,h_{n+t})=1$ for all $n\geq0$ and all $t\geq1$.
(Hint: Consider $h_{n+t}-2$.)

\item[(iii)]
Use (ii) to give a proof that there exist infinitely many prime
numbers.

\end{description}

(Hint: you may assume that every positive integer has a unique
factorisation into prime powers.)



ANSWER

\begin{description}

\item[(i)]
We have

\begin{eqnarray*}
h_0&=&2^{2^0}+1=2^1+1=3\\ h_1&=&2^{2^1}+1=2^2+1=5\\
h_2&=&2^{2^2}+1=2^4+1=17\\ h_3&=&2^{2^3}+1=2^8+1=257
\end{eqnarray*}

\item[(ii)]
We have

$$h_n+1-2=2^{2^{n+1}}+1-2=2^{2^n2^1}-2=(2^{2^n})^{2^t}-1=(h_n-1)^{2^t}-1$$

which is divisible by $h_n$, by the binomial theorem. Therefore
any common factor of $h_n$ and $h_{n+1}$ must divide 2. Since
$h_n$ is odd HCF$(h_n,h_{n+1})=2$ is impossible.

\item[(iii)]
Suppose that there are only finitely many distinct primes,
$p_1,p_2,\ldots p_k$. Let $P_n$ denote the set of primes which
appear to a strictly positive exponent in the prime power
factorisation of any one of $h_0,h_1\ldots,h_n$. By (ii) no
element of $P_m$ can appear in the prime factorisation of
$h_{m+1}$ so that $|P_n|\geq n$ which is impossible for $n>k$.

\end{description}




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