為何勾股數組都至少有一個為5的倍數？

相信初中的時候大家都學過畢氏定理，設a, b, c為直角三角形的對邊、鄰邊及斜邊，那a^2 + b^2 = c^2。符合a^2 + b^2 = c^2的一組正整數，我們稱之為勾股數組 (Pythagorean triple)，如 (3, 4, 5), (5, 12, 13), (8, 15, 17)...

大家有沒有發現每組勾股數組都至少有一個5的倍數？究竟為何如此？史丹福在數論書看過這個問題，覺得很有趣，自己想了一會，發現其實並不困難。證明如下：

For any positive integer x,

If x ≡ 1 (mod 5), then x^2 ≡ 1 (mod 5)

If x ≡ 2 (mod 5), then x^2 ≡ 4 (mod 5)

If x ≡ 3 (mod 5), then x^2 ≡ 9 ≡ 4 (mod 5)

If x ≡ 4 (mod 5), then x^2 ≡16 ≡ 1 (mod 5)

Therefore for all positive integer x, if x is not a multiple of 5, x^2 ≡ 1 or 4 (mod 5)

Let a, b, c be positive integers such that a^2 + b^2 = c^2.

Assume a, b, c are all not multiples of 5, then a^2, b^2, c^2 ≡ 1 or 4 (mod 5)

Case 1: a^2 ≡ 1 (mod 5), b^2 ≡ 1 (mod 5)

Then c^2 = a^2 + b^2 ≡ 1 + 1 = 2 (mod 5) ------ rejected as c^2 ≡/≡ 1 or 4 (mod 5)

Case 2: a^2 ≡ 1 (mod 5), b^2 ≡ 4 (mod 5)

Then c^2 = a^2 + b^2 ≡ 1 + 4 = 5 ≡ 0 (mod 5) ------ rejected as c^2 ≡/≡ 1 or 4 (mod 5)

Case 3: a^2 ≡ 4 (mod 5), b^2 ≡ 1 (mod 5)

Then c^2 = a^2 + b^2 ≡ 4 + 1 = 5 ≡ 0 (mod 5) ------ rejected as c^2 ≡/≡ 1 or 4 (mod 5)

Case 4: a^2 ≡ 4 (mod 5), b^2 ≡ 4 (mod 5)

Then c^2 = a^2 + b^2 ≡ 4 + 4 = 8 ≡ 3 (mod 5) ------ rejected as c^2 ≡/≡ 1 or 4 (mod 5)

There is contradiction. Therefore at least 1 of a, b, c is a multiple of 5.

QED