250 Problems in Elementary Number Theory - Sierpinski (1970)

SOLUTIONS 3333. The condition (x, y) = 1 is necessary since, for instance, 15 2 +202= 5\ while 7,t 15·20. Now, if (x, y) = 1 and x, y, z are positive integerssuch that r+y2 = z\ then, as we know from the theory of Pythagoreanequation, there exist integers m and n such that for instance x = m2_n2,y =2mn, Z2 = m 2 +n 2 • Suppose that 7,ty; thus 7,tm and 7 ,tn. It is easy tosee that the square of an integer not divisible by 7 gives, upon dividing by 7,the remainders 1, 2 or 4. Since 1 +2, 1 +4, and 2+4 cannot be such remainders,neither they are divisible by 7, it follows from equation z2 = m2++n2 that the numbers m and n must give the same remainders upon dividingby 7. Thus 71x = m 2 -w.34. The square of an integer not divisible by 7 gives upon dividing by7 the remainder 1, 2, or 4, hence the sum of such squares gives the remainder1, 2, 3, 4, 5, or 6. Thus, if a and b are integers such that 71a2+b 2 , then oneof them, hence also the other, must be divisible by 7.35*. The numbers x = 36k+14,y = (12k+5) (18k+7), k = 0, 1,2, ... ,have the desired property.In fact, we have obviously x(x+ 1)ly(Y+ 1) sincex(x+l) = 2·3(12k+5) (18k+7) = 6y,while 61y+ 1.The number y is not divisible by x since y is odd, while x is even. Thenumber y is not divisible by x+ 1 since 3jx+ 1, while 3,t y. The numbery+1 is not divisible by x since 18k+71x and 18k+7Iy, hence 18k+7 ,ty+1.Finally, the number y+l is not divisible by x+1 since 12k+5Ix+l and12k +5Iy, hence 12k+5,t y+ 1.For k ± 0, we obtain x = 14, y = 35, and it is easy to show that thereare no smaller numbers with the required property.36. For s < 10, we have of course ns = s. Next, studying successivemultiples of s, we obtain njO = 190, nll = 209, n12 = 48, n13 . 247, n14= 266, nlS = 155, n16 = 448, n17 = 476, n18 = 198, n19 = 874, n20 = 9920,n21 = 399, n22 = 2398, n23 = 1679, n24 = 888, n2S = 4975. Finally, we havenlOO = 19999999999900. In fact, two last digits of every number divisible by100 must be zero, and the sum of digits of every number smaller than199999999999 is obviously smaller than 100. See Kaprekar [11].37*. Let s be a positive integer, s = 2 1J 5{1t, where IX and (J are integers;;::: 0, and t is a positive integer not divisible by 2 or 5. By Euler's theoremwe have l()9'