本來想採用窮舉試探的方式來做這個演算法，後來發現還是有點麻煩。從網路上找來了一種求解方法如下：
當a為大於1的奇數2n+1時，b=2n^2+2n,c=2n^2+2n+1。實際上就是把a的平方數拆成兩個連續自然數。
編寫**如下：
#!/usr/bin/python
for n in range(1,101):
a = 2 * n +1
b = 2 * (n** 2) + 2 * n
c = b + 1
# check theresult
if a ** 2 +b ** 2 == c ** 2:
print("a= %d, b = %d, c = %d" %(a,b,c))
程式執行結果：
a = 3, b = 4, c = 5
a = 5, b = 12, c = 13
a = 7, b = 24, c = 25
a = 9, b = 40, c = 41
a = 11, b = 60, c = 61
a = 13, b = 84, c = 85
a = 15, b = 112, c = 113
a = 17, b = 144, c = 145
a 19, b = 180, c = 181
a = 21, b = 220, c = 221
a = 23, b = 264, c = 265
a = 25, b = 312, c = 313
a = 27, b = 364, c = 365
a = 29, b = 420, c = 421
a = 31, b = 480, c = 481
a = 33, b = 544, c = 545
a = 35, b = 612, c = 613
a = 37, b = 684, c = 685
a = 39, b = 760, c = 761
a = 41, b = 840, c = 841
a = 43, b = 924, c = 925
a = 45, b = 1012, c = 1013
a = 47, b = 1104, c = 1105
a = 49, b = 1200, c = 1201
a = 51, b = 1300, c = 1301
a = 53, b = 14程式設計客棧04, c = 1405
a = 55, b = 1512, c = 1513
a = 57, b = 1624, aqsczpxqbkc = 1625
a = 59, b = 1740, c = 1741
a = 61, b = 1860, c = 1861
a = 63, b = 1984, c = 1985
a = 65, b = 2112, c = 2113
a = 67, b = 2244, c = 2245
a = 69, b = 2380, c = 2381
a = 71, b = 2520, c = 2521
a = 73, b = 2664, c = 2665
a = 75, b = 2812, c = 2813
a = 77, b = 2964, c = 2965
a = 79, b = 3120, c = 3121
a = 81, b = 3280, c = 3281
a = 83, b = 3444, c = 3445
a = 85, b = 3612, c = 3613
a = 87, b = 3784, c = 3785
a = 89, b = 3960, c = 3961
a = 91, b = 4140, c = 4141
a = 93, b = 4324, c = 4325
a = 95, b = 4512, c = 4513
a = 97, b = 4704, c = 4705
a = 99, b = 4900, c = 4901
a = 101, b = 5100, c = 5101
a = 103, b = 5304, c = 5305
a = 105, b = 5512, c = 5513
a = 107, b = 5724, c = 5725
a = 109, b = 5940, c = 5941
a = 111, b = 6160, c = 6161
a = 113, b = 6384, c程式設計客棧 = 6385
a = 115, b = 6612, c = 6613
a = 117, b = 6844, c = 6845
a = 119, b = 7080, c = 7081
a = 121, b = 7320, c = 7321
a = 123, b = 7564, c = 7565
a = 125, b = 7812, c = 7813
a = 127, b = 8064, c = 8065
a = 129, b = 8320, c = 8321
a = 131, b = 8580, c = 8581
a = 133, b = 8844, c = 8845
a = 135, b = 9112, c = 9113
a = 137, b = 9384, c = 9385
a = 139, b = 9660, c = 9661
a = 141, b = 9940, c = 9941
a = 143, b = 10224, c = 10225
a = 145, b = 10512, c = 10513
a = 147, b = 10804, c = 10805
a = 149, b = 11100, c = 11101
a = 151, b = 11400, c = 11401
a = 153, b = 11704, c = 11705
a = 155, b = 12012, c = 12013
a = 157, b = 12324, c = 12325
a = 159, b = 12640, c = 12641
a = 161, b = 12960, c = 12961
a = 163, b = 13284, c = 13285
a = 165, b = 13612, c = 13613
a aqsczpxqbk= 167, b = 13944, c = 13945
a = 169, b = 14280, c = 14281
a = 171, b = 14620, c = 14621
a = 173, b = 14964, c = 14965
a = 175, b = 15312, c = 15313
a = 177, b = 15664, c = 15665
a = 179, b = 16020, c = 16021
a = 181, b = 16380, c = 16381
a = 183, b = 16744, c = 16745
a = 185, b = 17112, c = 17113
a = 187, b = 17484, c = 17485
a = 189, b = 17860, c = 17861
a = 191, b = 18240, c = 18241
a = 193, b = 18624, c = 18625
a = 195, b = 19012, c = 19013
a = 197, b = 19404, c = 19405
a = 199, b = 19800, c = 19801
a = 201, b = 20200, c = 20201
由於程式中加入了是否為勾股數的判斷，因此這個清單應該是準確的。這個小題目求解下來，我自己的感覺是做事的方法中演算法還是至關重要的！
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