1 毕达哥拉斯三元组的变体

(a) 找出 x2 + y2 = n 的所有解，其中 n 是 1 到 100 之间的整数。计算解为

( 1)2 + 22 = 5 都是 n = 5 的不同解。提示：有一个聪明的方法可以做到这一切

(b) 将 f(n) 写为 n 的此类解的数量。你能观察到任何模式吗？能

(c) 在不明确计算这些数字的情况下，你对下列数字有什么看法

^ 3125 = 55
. 3999 = 3 31 43
^ 4001，这是质数
. 4003，也是质数
^ 4096 = 212
. 5525 = 52 13 17

a2 + 4b4 = c4。这个结果清楚地暗示了费​​马大定理的 n = 4 情况。

(a) 假设 x、y 和 z 都是正的、成对互质的，并且满足 x4 + y4 = z2。使用
(x2; y2; z) 是勾股三元组的事实，构造一个整数 (a; b; c) 的三元组，使得
a2 + 4b4 = c4。

1 Variations on Pythagorean Triples
Problem 1. This problem will do some more investigation into numbers of the form x2 + y2.
(a) Find all solutions to x2 + y2 = n for n an integer between 1 and 100. Count solutions as
the same only if they are literally the same, so one has that 12 + 22 = 5, 22 + 12 = 5, and
( 1)2 + 22 = 5 are all di erent solutions for n = 5. Hint : there is a clever way to do this all
at once.
(b) Write f(n) to be the number of such solutions for n. Can you observe any patterns? Can
you prove any of your observations?
(c) Without explicitly computing these numbers, what do you think f of the following numbers
is?
 3125 = 55
 3999 = 3  31  43
 4001, which is prime
 4003, which is also prime
 4096 = 212
 5525 = 52  13  17
Problem 2. This problem will walk you through a proof that there are no solutions to x4+y4 = z2 other than the trivial ones. Along the way, you will also show that there are no solutions to a2 + 4b4 = c4. This result clearly implies the n = 4 case of Fermat’s last theorem.
Let S1 = f(x; y; z)jx; y; and z are positive, pairwise coprime, and x4 + y4 = z2g and S2 be the same for the equation a2 + 4b4 = c4.
(a) Assume that x, y, and z are all positive, pairwise coprime, and satisfy x4 + y4 = z2. Using the fact that (x2; y2; z) is a Pythagorean triple, construct a triple of integers (a; b; c) such that a2 + 4b4 = c4.