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In which the product is assumed to be inﬁnite. In order to see what kind of series will result, I multiplied actually a great number of factors and found 1 − x − x2 + x5 + x7 − x12 − x15 + x22 + x26 − x35 − x40 + . . The exponents of x are the same which enter into the above formula; 1 also the signs + and − arise twice in succession. It suﬃces to undertake this multiplication and to continue it as far as it is deemed proper to become convinced of the truth of these series. Yet I have no other evidence for this, except a long induction which I have carried out so far that I cannot in any way doubt the law governing the formation of these terms and their exponents.

We shall say that a rational number r = n assumed irreducible) is divisible by pk , if m is divisible by pk and n is not divisible by p. For rational numbers r, s, the congruence r ≡ s mod pk means that r − s is 1 divisible by pk . ) These congruences possess the usual 5 properties of congruences: if r ≡ s mod pk and s ≡ t mod pk , then r ≡ t mod pk ; if r ≡ s mod pk and the denominator of t is not divisible by t, then rt ≡ st mod pk ; etc. 9. For a prime p ≥ 5, 1+ 1 1 + ···+ 2 p−1 is divisible by p2 .

Mk−1 ) and if m1 = s(m1 , . . , mk−1 )+1 = (k−1)+1, then, on one hand, m1 = n1 +1, and on the other hand, n1 + 1 = k, hence mi = ni+1 + 1, if i ≥ k − n1 = 1, hence m1 = n2 + 1; this is not possible, since n2 > n1 . The fact that the above transformation is 1 − 1 follows from the existence of an inverse transformation: s consecutive numbers m1 . . . . . . mk−1 s → m1 . . . . . mk−1 −1 · · · − 1 ⏐ ↓ ... ↓ ⏐ s ←− 1 . . 1 → s m1 . . mk−s . . mk−1 −1 . . − 1 LECTURE 3. COLLECTING LIKE TERMS AND MISSED OPPORTUNITIES 47 (that is, we subtract 1 from each of the s consecutive numbers in the right end, collect these ones into one number s and place this s before m1 ) or, in formulas: ⎧ if i = 1, ⎨ s, mi−1 , if 2 ≤ i ≤ k − s, (m1 , .