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1 : 11 x) log x 39 Combining such an estimate with knowledge of a speci c x0 and computations verifying Bertrand's Hypothesis for x < x0 , a proof of Bertrand's Hypothesis follows. Similar work by others has been obtained. In particular, Ramanujan gave an argument for Bertrand's Hypothesis and noted that there are at least 5 primes in (x; 2x] for x 20:5. Our next theorem is a variation of Chebyshev's Theorem. The proof below is due to Erd}os. Theorem 33. If n is a su ciently large positive integer, then 1 n < (n) < 3 n : 6 log n log n Proof.

We used that X x X x ? X x + : A(z; x) = x] ? + p z p p1 z . If r = 0, then clearly (n) = 1. If r > 0, then r = (1 ?

By (i) and (ii), there exist exactly three incongruent solutions to f (x) 0 (mod 33 ) given by 7, 16, and 25. Observe that solving f (x) 0 (mod 3e ) is actually easy since f (x) = (x + 2)2 . If k is the least integer greater than or equal to e=2, then f (x) 0 (mod 3e ) if and only if x + 2 0 (mod 3k ). k solutions given by 3k ` ? k g. A third example. Here we calculate all incongruent solutions modulo 175 to x3 + 2x2 + 2x ? 6 0 (mod 175): Since 175 = 52 7, we consider f (x) 0 (mod 25) and f (x) 0 (mod 7) where f (x) = x3 + 2x2 + 2x ?

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