250 Problems in Elementary Number Theory - Sierpinski (1970)

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120 250 PROBLEMS IN NUMBER THEORYOn the other hand,and consequently,yD+aoXl-at = 4mn+l2m = YD- [(4mn+ l)m-n]4mn+l '(4mn+l) [VD+(4mn+l)m-n]D- [(4mn+l)m-n]ZWe easily check thatwhich yieldsand since ao < yD < ao+l, orxl = VD+(4mn+l)m-n4mn+lwe get(4mn+l)m+n < VD < (4mn+l)m+n+l,12m < Xz < 2m+ 4mn+lConsequently, the integral part of Xz equals az = 2m. We have, therefore,X2 = az+ l/x3> which gives X3 = 1/(x2-a2). However,yD+(4mn+l)m-n 2 _ Y:D-(4mn+l)m-nX2-aZ = 4mn+l m - 4mn+l .Consequently, we have- (4mn+l) [vD+(4mn+l)m+n] _ :ID (4 +1) + - 'D+X3 - D-[(4mn+l)m+n]Z - J' + mn m n - V aowhich implies that the integral part of X3 is 2ao, and that the number VI>has the expansion into the arithmetic continued fraction with the threetermperiod, formed of numbers 2m, 2m and 2ao.One can show that all positive integers D for which the exREMARK.pansion of y D into arithmetic continued fraction has a three:-term periodare just the above considered numbers D. See Sierpinski [32].
SOLUTIONS 12124S. Computing the values of functions cp(n) and den) for n ~ 30 fromthe well-known formulae for these functions. i.e. if n = qf'q~2 ... q~', thenwe easily see that the only values n ~ 30 for which cp(n) = den) are n = 1, 3,S, 10, IS, 24, and 30. We have here cp(I) = d(l) = 1, cp(3) = d(3) = 2, cp(8)= deS) = 4, cp(lO) = d(lO) = 4, cp(IS) = d(IS) = 6, cp(24) = d(24) = 8,cp(30) = d(30) = 8.REMARK.It was proved that there are no other solutions of the equationcp(n) = den) in positive integers n. It can be shown that for n > 30 we havecp(n) > den); see P6lya and Szego [15, Section VIII, problem 45].249. We easily check that for positive integer k and integer s;?: 0 wehave(I.!) ( _1 ) ( _1 ) _ s+ 1+ k 1+ k+l ... 1+ k+s - 1+ k . (1)A positive rational number w-l can be always represented in the form w-1 = min where m and n are positive integers (not necessarily relativelyprime) and n > g. It suffices to take k = nand s = m-l; then the righthandside of (1) will be equal to w. In this way we obtain the desired decompositionfor the number w.250* . We shall first prove that every integer k ;?: 0 can be in at least oneway represented in the formwhere m is a positive integer, and the signs + and -(1)are suitably chosen. Theassertion holds for the number 0 since 0 = 12+22_32+42_52_62+72. It isalso true for the numbers 1, 2, and 3 since 1 = 12, 2 = _12_22_32+42,3 = -12+22,4 = _12_22+32.Now, it suffices to prove that our theorem is true for every positive integerk, and since it is true for numbers 0, 1, 2, and 3, it suffices to prove that ifthe theorem is true for an integer k ;?: 0, it is also true for the number k+4.Suppose, then, that the theorem is true for the number k; thus, there exists
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250 PROBLEMSIN ELEMENTARY NUMBER TH
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PROBLEMS 553. Prove that for every
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PROBLEMS 777. Prove that every prim
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PROBLIMS 9contains at least one pri
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PROBLEMS 11128*. From a particular
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•• oaLEHS13153. Prove that the
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PROBLEMS 15167*. Prove that for eve
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PROBLEMS 17189. Using the identity(
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PROBLEMS 19209*. Prove that the sum
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PRO.LlMS 21positive integers which
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SOLUTIONSI. DIVISIBILITY OF NUMBERS
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SOLUTIONS 2510. These are all odd n
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SOLUTIONS 2716. We have 312 3 +1, a
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SOLUTIONS 29Since HI = 2"+2 > H, th
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SOLUTIONS 31Since 2 3 - 3 (mod 5) a
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SOLUTIONS 3333. The condition (x, y
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SOLUTIONS3S'2, '3, while n > 3, the
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SOLUTIONS 37a, b, c give three diff
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SOLUTIONS 39hence a+b+c ~ 5+7+9 ~ 2
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SOLUTIONS 41rectangular triangle wi
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SOLUTIONS 43= 1. If we had plQ, the
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SOLUTIONS 4566*. The progression ll
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SOLUTIONS 47primes, then the differ
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SOLUTIONS 49have, for example, 52 =
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SOLUTIONS 51The least such number i
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SOLUTIONS 53n 2 -1 is a product of
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SOLUTIONSssnumbers are consecutive
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SOLUTIONS 57The problem arises whet
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SOLunONSS9which implies that 2 N /p
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SOLUTIONS 61follows that we must ha
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SOLUTIONS 63to note that (for k = 1
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SOLUTIONS 65(where p > F4)' Let us
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SOLUTIONS 67171 (34k+2)22 +1, 171 (
Page 81 and 82: SOLUTIONS 69Obviously, gk(X) is a p
Page 83 and 84: SOLUTIONS 7130t+r where t is an int
Page 85 and 86: SOLUTIONS 73136. We easily find tha
Page 87 and 88: SOLUTIONS 75145. Our equation is eq
Page 89 and 90: SOLUTIONS 77If we had 161d, then by
Page 91 and 92: SOLUTIONS 79154*. LEMMA. If a, b, c
Page 93 and 94: SOLUTIONS 81157. Suppose that theor
Page 95 and 96: SOLUTIONS 83160. We must have x ::;
Page 97 and 98: SOLUTIONS 85f h Ill h' h' I' 2 1 6I
Page 99 and 100: SOLUTIONS 87For s = 3, the equation
Page 101 and 102: SOLUTIONS89We must therefore have X
Page 103 and 104: SOLUTIONS 91integer s at least one
Page 105 and 106: SOLUTIONS 93k is an integer> 3, the
Page 107 and 108: SOLUTIONS95REMARK.One can prove tha
Page 109 and 110: SOLUTIONS 97= 2 x , hence Y > 1, an
Page 111 and 112: SOLUTIONS 99then 2Zk(zz+l) = y3_1 =
Page 113 and 114: SOLUTIONS 101MISCELLANEA200. The eq
Page 115 and 116: SOLUTIONS 103The assertion can be s
Page 117 and 118: SOLUTIONS 1052" == [(mod 2k), and c
Page 119 and 120: SOLUTIONS 107for instance [a] + I,
Page 121 and 122: SOLUTIONS 109225*. We shall prove b
Page 123 and 124: SOLUTIONS 111itive integers, as in
Page 125 and 126: SOLUTIONS113We may assume that u ~
Page 127 and 128: SOLUTIONS 115square. On the other h
Page 129 and 130: SOLUTIONS 117namely numbers 13 and
Page 131: SOLUTIONS 119For positive integers
Page 136 and 137: 124 REFERENCES24. W. Sierpmski, Sur
Page 138 and 139: The late Waclaw SierpiJiski, a memb
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250 Problems in Elementary Number Theory - Sierpinski (1970)

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