
Unit 3 Items to Support Formative Assessment
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Fermat Numbers in the Pascal Triangle
... which is impossible because q does not divide Fm . Hence, every prime divisor of d divides k as well. To show that d|k, we need to show that if q α || d for some α ≥ 1, then q α |k. Assuming that this were not so, it follows that q β || k for some β < α. But in this case, nβ = 0 and kβ 6= 0 which, v ...
... which is impossible because q does not divide Fm . Hence, every prime divisor of d divides k as well. To show that d|k, we need to show that if q α || d for some α ≥ 1, then q α |k. Assuming that this were not so, it follows that q β || k for some β < α. But in this case, nβ = 0 and kβ 6= 0 which, v ...