250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 31Since 2 3 - 3 (mod 5) and 24 = 3 (mod 13), we get 2 4k + 3 == 3 (mod 5) and2121+4 == 3 (mod 13) for k = 0, 1, 2, .... Therefore 512 4ṭ + 3 -3 and 13/212"+4_3-for k = 0, 1, 2, ....Next, 2 6 == -1 (mod 65), which implies that 212 == 1 (mod 65) and therefore2"+ 12 - 3 == 2"-3 (mod 65), which shows that the sequence of remaindersmodulo 65 of the sequence 2"-3 (n = 2, 3, ...) is periodic with period 12.To prove that none of the numbers 2"-3 (n = 2, 3, ...) is divisible by 65it is sufficient to check whether the numbers 2~-3 for n = 2, 3, ... , 13 aredivisible by' 65. We find easily that the remainders upon dividing by 65 areI, 5, 13,29,61,60, 58, 54,46,30,63,64, and none of these remainders is zero.27*. It is known (see, for instance, Sierpiilski, [37, p. 215]) that thefour smallest composite numbers n, such that nI2"-2, are 341, 561, 645,and 1105. For 341, we have 341,¥3 341 -3 since, by Fermat's theorem, 3 30 ,= 1 (mod 31), which implies 3330 = 1 (mod 31), hence 3 341 = 3 11 (mod 31).In view of 3 3 == -4 (mod 31), we get 3 9 == -64 = -2 (mod 31), hence3 11 = -18 (mod 31). Therefore 3 341 _3 = 3 11 -3 = -21 (mod 31), and31 ,r 3 341 _ 3, which implies 341 = 11 · 31 ,r 3 341 _ 3. On the other hand, 561= 3 ·11 · 1713 561 -3 since 1113 1 °-1 which implies 1113 39 °-1 and 11/3 341 -3,and also 1713 16 -1 which implies 17\3 16 • 3s -1 = 356°_1. Thus 1713 s61 -3.Thus, the least composite number n such- that n/2"-2 and nI3"-3 isthe number n = 561.The number 645 is not a divisor of 3645_3 since 645 = 3 · 5 · 43, while3 42 = 1 (mod 43) which implies 3 42 . 15 == 1 (mod 43). Thus 3 630 = 1 (mod 43),and 3645 == 3 1S (mod 43). Since 3 4 = -5 (mod 43), we have3 6 = -45 = -2 (mod 43), 3 12 = 4 (mod 43), 3 15 == 108 = 22 (mod 43).Therefore 3645_3 = 19(mod 43), which implies 43,¥ 3645_3.Onthe other hand, we have 1105\3 1105 -3. Indeed, 1105 = 5· 13 · 17, 3 4== 1 (~od 5), and 3 1104 == 1 (mod 5), and 513 1105 -3. Next, 3 12 = 1 (mod 13),3 1104 = 1 (mod 13) and 1313 1105 _3. Finally, 3 16 == 1 (mod 17), and since1104 = 16 · 69, we get 3 1104 = 1 (mod 17), which implies 1713 1105 -3.Thus, two smallest composite numbers for which n\2"-2 and nI3"-3 are561 and 1105.REMARK. We do not know whether there exist infinitely many compositenumbers n for which n\2"-2 and n)3"-3. This assertion would follow froma conjecture of A. Schinzel concerning prime numbers ([22]). For prime numbersn, both relations nI2"-2 and nI3"-3 hold because of Fermat's theorem.