By Leo Moser

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**Example text**

One separates the numbers 1, 2, . . , p−1 into 2 classes, residues and nonresidues. If p is large enough one of these classes will contain, by Van der Waerden’s theorem, 49 terms in arithmetic progression, say a, a + b, a + 2b, . . , a + 48b. Now if ab = c then we have 49 consecutive numbers of the same quadratic character, namely c, c + 1, c + 2, . . , c + 48. If these are residues we are done. If nonresidues then suppose d is the smallest nonresidue of p. If d ≥ 7 we are finished for then 1, 2, .

We next consider the composition of 2n as a product of primes. , 2n n pEp (n) . = p Clearly Ep (n) = ep (2n) − 2ep (n) = i 2n n −2 i i p p . Our alternative expression for ep (n) yields 2sp (n) − sp (2n) . p−1 Ep (n) = In the first expression each term in the sum can easily be seen to be 0 or 1 and the number of terms does not exceed the exponent of the highest power of p that does not exceed 2n. Hence Lemma 5. Ep (n) ≤ logp (2n). Lemma 6. The contribution of p to 2n n does not exceed 2n. The following three lemmas are immediate.

Suppose now that pq = λ is an approximation to λ. We may assume the approximation good enough to ensure that pq lies in the interval (λ − 1, λ + 1), 40 Chapter 4. Irrational Numbers y = f (x) f (p/q) λ p/q Figure 2 and is nearer to λ than any other root of f (x) = 0, so that f (p/q) = 0. Clearly (see Figure 2), f p q = 1 1 |a0 pn + a1 pn−1 q + · · · + an q n | ≥ n n q q and f (p/q)