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22 lines
402 B
22 lines
402 B
a = 1
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b = 1
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c = 1
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def is_pythag(a, b, c):
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return a < b < c and a**2 + b**2 == c**2
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for a in range(1, 1000):
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for b in range(1, 1000):
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for c in range(1, 1000):
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if c % 100 == 0:
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print(f"iter {a} {b} {c}", end="\r")
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if is_pythag(a, b, c) and a+b+c==1000:
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print(f"a {a} b {b} c {c}")
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else:
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continue
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