Solutions to PS 3

(Schumer 2.3 Prob. 1) Show that if n is a perfect square and a perfect cube, then it is a perfect sixth power.

Solution: We can factor n into primes,

$\begin{displaymath}n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}} \end{displaymath}$

where p_i are distinct prime factors. n is a perfect square implies $2 \vert a_{i} \forall i$ , and n is a perfect cube implies $3 \vert a_{i} \forall i$ . Thus $6 \vert a_{i} \forall i$ , and so nis a perfect sixth power.

2. (Schumer 2.3 Prob. 3 (b)) Prove that there are infinitely many primes of the form 6k+5.

Solution: Suppose not, and there are finitely many such primes, $\{p_{1}, p_{2}, \cdots , p_{N} \}$ . Consider the number $K=6p_{1} p_{2} \cdots p_{N} -1$ . $K \equiv 5 (mod 6)$ , and as such cannot be the product of only primes $\equiv 1 (mod 6)$ . Therefore, Kmust have a prime factor $p \equiv 5 (mod 6)$ . But $K \equiv -1 (mod p_{i}) \forall i$ , and so $p \ne p_{i}$ for any i, and so our list did not contain all the primes $\equiv 5 (mod 6)$ after all.

3. (Schumer 2.3 Prob. 8) If p and q are twin primes with 3 < p < q, then show that 9 divides pq+1 and that pq+1 is a perfect square. Recall that p, q are twin primes if q-p=2.

Solution: q=p+2, so pq+1=p(p+2)+1=(p+1)², so pq+1 is a perfect square. The possible residues of a prime p > 3 mod 9 are 1,2,4,5,7,8. If $p \equiv 1 (mod 9)$ , then $q \equiv 3 (mod 9)$ which is impossible, since q is supposed to be prime. Continuing in this fashion, we see that the only possibilities for (p,q) mod 9 are (2,4), (5,7), (8,1). In these $pq+1 \equiv 0 (mod 9)$ .

4. (Schumer 2.3 Prob. 14) Show that 3 is the only prime of the form k⁴+k²+1.

Solution: P(k)=k⁴+k²+1=(k⁴+2k²+1)-k²=(k²+1)²-k²=(k²+k+1)(k²-k+1). Thus if P(k) is prime for some $k \in mathbb{Z}$ , then one of the factors must be $\pm 1$ . Solving the equations $k^2 \pm k +1 = \pm 1$ shows that $k=0, \pm 1, pm 2$ . Checking P at all these values shows that the only prime value taken by Pis 3.

5. (Schumer 2.3 Prob. 15) We say that the real numbers $r_1, \cdots , r_n$ are linearly independent [over the integers] if the only solution to k₁*r₁+k₂*r₂+...+k_n*r_n=0 where k_i are integers is $k_1=k_2=\cdots=k_n=0$ . Show that if $p_1, p_2, \cdots , p_n$ are n distinct primes, then the real numbers $log(p_1), log(p_2), \cdots, log(p_n)$ are linearly independent.