Bewijs de contrapositieve variant: | ||
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Stel n = k2 , dan zijn er vier gevallen: | ||
1. | k (mod 4) = 0 Dan is k = 4q en k2 = (4q)2 = 16q2 = 4 • (4q2 ) dus is n (mod 4) = 0 |
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2. | k (mod 4) = 1 Dan is k = 4q + 1 en k2 = (4q + 1)2 = 16q2 + 8q + 1 = 4 • (4q2 + 2q) + 1 dus n (mod 4) = 1 |
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3. | k(mod 4) = 2 Dan is k = 4q + 2 en k2 = (4q + 2)2 = 16q2 + 16q + 4 = 4 • (4q2 + 4q + 1) dus n (mod 4) = 0 |
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4. | k (mod 4) = 3 Dan is k = 4q + 3 en k2 = (4q + 3)2 = 16q2 + 24q + 9 = 4 • (4q2 + 6q + 2) + 1 dus n (mod 4) = 1 |
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