3. Given that p does not divide n for all primes p less than the cube root of n, show that n>1 is either a prime or the product of two primes.

Proof: Suppose not. That is, suppose there is an integer n so that n>1, p does not divide n for any prime p less than the cube root of n, and n contains at least three prime factors. Then n = p₁p₂ p₃k, where p₁, p₂, and p₃ are distinct primes and k is an integer with k³ 1. Denote the cube root of n by cbrt(n). By assumption, since no prime less than or equal to cbrt(n) divides n, then p₁>cbrt(n), p₂>cbrt(n), and p₃>cbrt(n). Hence p₁p₂ p₃>( cbrt(n))³>n. But n = p₁p₂ p₃k, where k>1. So n>p₁p₂p₃. Thus n> p₁p₂ p₃>n, which is a contradiction. It follows that the supposition is false and the given statement is true.

2. If x, y, z is a primitive Pythagorean triple, prove that x+y and x-y are congruent modulo 8 to either 1 or 7.

Proof: By Theorem 11.1, there exist integers s and t such that s>t>0, s and t have opposite parity, x =2st, and y = s²-t². Since s and t have opposite parity, s+t is odd. Hence

(s+t)² º 1 (mod 8).* Thus

x + y = 2st + (s²- t²) = s² + 2st - t² = s² + 2st + t ² - 2t ² = (s+t)² - 2t² º 1 - 2t ² (mod 8).

If t is even, then 8 divides 2t², and so x + y º 1 - 2t ² º 1 (mod 8).

If t is odd, then t = 2n+1 for some integer n, and so

x + y º 1 - 2t² = 1 - 2(2n+1)² = 1 - 2(4n²+4n+1) = -8n² - 8n -1 º -1 º 7 (mod 8).

Thus x +y is congruent modulo 8 to either 1 or 7.

The proof that x - y is congruent modulo 8 to either 1 or 7 is almost identical.

*[Why? If k is an integer, then (2k +1)² = 4k²+4k+1 = 4k(k+1) + 1. But k(k+1) is even; say k(k+1) = 2m. Hence (2k+1)² = 4(2m)+1 = 8m + 1.]

1. Show that the equation x²+ y² = z³ has infinitely many solutions for x, y, z positive integers.

Proof: There are infinitely many integers that are greater than 3. For each such integer n, let x = n(n² - 3) and y = 3n² - 1. Then

Let z = n² + 1. Then x² + y² = z³.

3. In a Pythagorean triple x, y, z, prove that not more than one of x, y, or z can be a perfect square.

Proof: Let x, y, z be a Pythagorean triple.

Case 1: Suppose both x and y are perfect squares. Then x = u² and y = v² for some integers u and v. Hence x² + y² = (u^{2) 2} + (v^{2) 2} = u⁴ + v⁴ = z². But this contradicts theorem 11.3. Thus case 1 cannot occur.

Case 2: Suppose both x and z are perfect squares. Then x = u² and z = v² for some integers u and v. Hence x² + y² = (u^{2) 2} + y² = z² = (v^{2) 2}. So u⁴ + y² = v⁴ , and thus y² = v⁴ - u⁴. But this contradicts theorem 11.4. Thus case 2 cannot occur.

Case 3: Suppose both y and z are perfect squares. This case is virtually identical to case 1 and also cannot occur.

Therefore, not more than one of x, y, or z can be a perfect square.