Math 514 homework #3 Due date: 19950906 D. J. Bernstein 19950828 Feel free to use the following two theorems in the subsequent exercise. Theorem. Let x, y, n be integers with y^{2} + 19 = x^{n} and n ≥ 3. Then x is odd and y is not divisible by 19. Proof. If x is even then 2^{n} divides x^{n} = y^{2} + 19 so (since n ≥ 3) 8 divides y^{2} + 19. Thus y is odd. If y is of the form 4k + 1 then y^{2} + 19 = (4k + 1)^{2} + 19 = 16k^{2} + 8k + 20; if y is of the form 4k + 3 then y^{2} + 19 = (4k + 3)^{2} + 19 = 16k^{2} + 24k + 28. Either way, y^{2} + 19 is not divisible by 8; contradiction. If y = 19k then (19k)^{2} + 19 = x^{n}, so 19 divides x^{n}, so (since 19 is prime) 19 divides x. Write x = 19m. Then 19^{2}k^{2} + 19 = 19^{n}m^{n}, so 19k^{2} + 1 = 19^{n−1}m^{n}, so 1 = 19^{n−1}m^{n} − 19k^{2}, so (since n ≥ 2) 1 is a multiple of 19; contradiction. Theorem. Let x, y, n be integers with y^{2} + 19 = x^{n} and n ≥ 3. Then y + \sqrt{−19} and y − \sqrt{−19} are coprime in the ring R = Z[\sqrt{−19}]; i.e., the ideal I = (y + \sqrt{−19})R + (y − \sqrt{−19})R is equal to R. Proof. We know y is not divisible by 19, so (since 19 is prime) y is coprime to 19 in Z. Hence 19a + yb = 1 for some integers a, b. Similarly, we know x^{n} is odd, so 1 = 2c + x^{n}d for some integers c, d. By hypothesis y + \sqrt{−19} ∈ I and y − \sqrt{−19} ∈ I. Thus 2y = (y + \sqrt{−19}) + (y − \sqrt{−19}) ∈ I and 2\sqrt{−19} = (y + \sqrt{−19}) − (y − \sqrt{−19}) ∈ I. Next 38 = −\sqrt{−19}(2\sqrt{−19}) ∈ I. Thus 2 = a(38) + b(2y) ∈ I. Finally, x^{n} = y^{2} + 19 = (y + \sqrt{−19})(y − \sqrt{−19}) ∈ I, so 1 = c(2) + d(x^{n}) ∈ I. Exercise 8. Assuming that Pic Z[\sqrt{−19}] is a group of size 3, find all integer solutions to y^{2} + 19 = x^{5}. Exercise 6 (clarification). Assume that p is prime.