Proposed by Jean-Marie De Koninck, Université Laval, Québec, Canada. (a) Let φ denote the Euler φ function, and let γ(n) = p n

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1 Proposed by Jean-Marie De Koninck, Université aval, Québec, Canada. (a) et φ denote the Euler φ function, and let γ(n) = p n p, with γ(1) = 1. Prove that there are exactly six positive integers such that φ(n) = (γ(n)) 2. (b)* et σ(n) denote the sum of divisors of n. Prove or disprove that the only solutions to σ(n) = (γ(n)) 2 are n = 1 and n = Solution: The first result will follow from a more general theorem, which will be proved. This theorem includes an algorithm whose output proves (a). There will also be an outline of a possible resolution of (b); some partial results will be presented. Theorem. For any integer k > 0, there are a finite number of integers n such that φ(n) = γ(n) k. Furthermore, there is a sophisticated algorithm to find all such n. Proof of the theorem: Write γ = γ(n) and φ = φ(n), and suppose γ k = φ. Write n as a product of prime numbers: n = p αi i, so that γ = p i and φ = (p i 1)p αi 1 i, where p 1 < p 2 < < p, and α i > 0, for all applicable i. The finiteness result will follow after three claims are proven: (1) α i k + 1, for all i = 1, 2,...,. If φ = γ k, then equating the above formulas for γ and φ implies that (p i 1) = φ p 1 αi i = γ k p 1 αi i = p k+1 αi i. (E1) Since the left-hand side is an integer, the right-hand side must also be one. Hence k + 1 α i 0, which proves (1). (2) k + 1. First, p 1 = 2; this can be shown with (E1): if n is odd, then the left-hand side of (E1) is even, and the right-hand side is odd. Now we look for the highest power of 2 which divides into both sides. That power is at least on the left-hand side, and on the right-hand side it is k + 1 α 1 k + 1. This proves (2). (3) p Q k () Q k (k + 1), where Q k (h) is the function defined by Q k (h) = { 2, if i = 1; 1 + Q k (1) k Q k (2) k Q k (i 1) k, otherwise. The second inequality in the statement of (3) follows because Q k is an increasing function. To prove the first inequality is true, we will show R(i) is true for j =, where R(i) denotes the proposition For all 1 i j, p i Q k (h), and there exist j integers β 1, β 2,..., β j between 0 and k + 1 such that p β1 1 pβ2 2 pβj j p k+1 αj+1 j+2 p k+1 α = (p i 1). (E2) i=j+1 This will be proven by induction on j. The proposition R(1) is true because p 1 = 2, and the equality (E2) is the equality (E1) above when j = 1. Now suppose R(j) is true, and consider the prime factors of p j+1 1. Since p j+1 1 is strictly less than p j+1,..., p, the only possible prime factors of p j+1 1 are among p 1, p 2,..., p j. 1

2 (Otherwise some prime divides evenly into the right-hand side of (E2) but not the left-hand side.) So we may write p j+1 1 = p δ1 1 pδ2 2 pδj j, for some δ 1, δ 2,..., δ j. The highest power of p i that divides evenly into the left-hand side of (E2) is β 1 (if i j), so 0 δ i β i, for all i j. Thus p j+1 = 1 + p δ1 1 pδ2 2 pδj j p β1 1 pβ2 2 pβj j To prove (E2) for j + 1, note that i=j+2 p k+1 1 p k+1 2 p k+1 j 1 + Q k (1) k Q k (2) k Q k (i) k = Q k (j + 1). (p i 1) = i=j+1 (p i 1) p j+1 1 = pβ1 1 pβ2 2 pβj j p k+1 αj+1 j+2 p k+1 α p δ1 1 pδ2 2 pδj j = p β1 δ1 1 p β2 δ2 2 p βj δj j p k+1 αj+1 j+2 p k+1 α, so R(j + 1) is true, if R(j) is. This proves claim (3). Putting claims (1), (2), and (3) together, we get n = p αi i p k+1 Q k (k + 1) k+1 [Q k (k + 1) k+1 ] [Q k (k + 1)] (k+1)2, which shows that there are only a finite number of integers n such that φ = γ k. Now for the algorithm. A non-sophisticated algorithm exists to determine all values of n such that φ = γ k, namely the one which checks all integers between 1 and [Q k (k +1)] (k+1)2. A more sophisticated one exists, which does not require (nearly) as much time. This algorithm proceeds along the line of the proof, and proceeds in two phases. Quantities in brackets should be read as arrays (ordered lists). (I) Put (j, 1, [k + 1], [2]) on a stack (a queue also works), for all j between 1 and k + 1. (II) While there is an item (, j, [β 1, β 2,..., β j ], [p 1, p 2,... p j ]) on the stack, do the following: If j =, then output the factorization p β1 1 pβ2 2 pβj j (as a value of n where φ(n) = γ(n) k ). Otherwise, find all sequences [δ 1, δ 2,..., δ j ] such that (a) β i 1 δ i 0, for i = 1, 2,..., j; and (b) P = 1 + p δ1 1 pδ2 2 pδj j is prime and greater than p j ; and put the item (, j + 1, [β 1 δ 1, β 2 δ 2,..., β j δ j, k + 1], [p 1, p 2,..., p j, P ]) on the stack. Note that the item (, j, [β 1, β 2,..., β j ], [p 1, p 2,, p j ]) being on the stack means that (E2) is true. This proves the theorem. Using the algorithm given in the theorem for k = 2 produces the numbers 1 (where = 0), 2 3 ( = 1), and ( = 2), and and ( = 3). (The bound given by the theorem (101 9 ) is many orders of magnitude larger than the actual upper bound (43,434), but this is typical of finiteness results.) 2

3 Turning to part (b), a similar approach was attempted, and some partial results have been obtained. Using the prime factorization of n, we have σ = σ(n) = S 1 S 2 S, where S i = 1 + p i + p 2 i + + pαi i. The goal then is to find the prime factorization of S i, for all i, assuming that σn = γ(n) 2. This means that the prime factors of S i are among p 1, p 2,... p. Also, no cube of p j divides into S i, so this suggests a combinatorial ( finite ) approach may also work here. Partial results are given below as a list of lemmas. emma 1. Suppose σ = γ 2, and n > 1. Then: (a) There is an I such that α I = 1; (b) p 1 = 2; (c) p i S i ; (d) p j S I, if j > I; (e) If p 1 S i, then α i is even, for any i > 1; (f) If p 1 S i, then α i 1 (mod 4), for any i; (g) If i = max{j : p 1 S j }, then α i = 1. (h) S i 1, and if S i = p j, then p i < p j (and hence i < j); (i) If S j = p 2 i and i < j, then (i, j) = (1, 2), p 2 = 3, and α 2 = 1. Proof: If α i 2 for all i, then S i 1 + p i + p 2 i > p2 i, and thus σ = S i > p 2 i = γ 2. This proves (a). To prove (b), suppose that every p i is odd. Part (a) implies that p βi i = S I = 1 + p I, for some nonnegative integers β 1,..., β. But the left-hand side is the product of odd numbers and is hence odd; the right-hand side is even. This contradiction proves (b). Part (c) follows because S i 1 (mod p i ) (since p j i 0 (mod p i), for all i and j). Part (d) follows because otherwise p j p βi i = S I = 1 + p I < 1 + (p j 1) = p j, where β i are nonnegative integers and β j 1. To prove (e), suppose p 1 S i. Then S 1 is odd, and since i > 1, p i > 2, and p i is odd as well. Thus, 1 S i 1 + p i + p 2 i + + p αi i α i + 1 (mod 2), which proves (e). A similar equality shows that if p 1 S i, then α i is odd, which proves part of (f). Now, if α i = 4m + 3 for some integer m, then ( ) S i = 1 + p i + p 2 i + + p 4m+3 i = (1 + p i ) 1 + p 2 i + p 4 i + p 2(2m+1) i, which implies that p 2 1 S i (because 1+p 2 i +p4 i +p2(2m+1) i is even). Thus, for every j i, p 1 S j, and hence α j 2, by (e). By hypothesis, α i 3; but then these inequalities violate (a). This proves (f). If (g) does not hold, then α i 5, by (f). If p 2 1 S i, then p 4 k > p2 k p 1, if k = i; S k > p k, if k = 1; p 2 k, if k 1, i, since α k is even. 3

4 Multiplying these inequalities together for all k results in the inequality σ > γ 2. If p 1 S j, where j < i, then p 5 k > p2 k p 1 p j, if k = i; p S k > k, if k = j; p 1, if k = 1;, if k 1, i, j. p 2 k Once again, multiplying these inequalties together yields σ > γ 2. This proves (g). If (h) does not hold, then 1 + p j < 1 + p i = S i = p j. Finally, we show (i): Note that p αj j < S j = p 2 i < p2 j, so 1 α j < 2. Hence α j = 1, which implies that p 1 S j by (e). Hence 1 + p j = S j = p 2 1 = 4, or p j = 3. But then we must have j = 2 as well, since p 3 5. Now for the constructive results: emma 2. If = 1 or 2, then σ(n) γ(n) 2 (where and α come from the prime factorization of n). Proof: If = 1, then emma 1(a) and (b) imply that n = 2 1, but σ(2) = If = 2, then emma 1(c) implies that p 2 1 = S 2 and p 2 2 = S 1. emma 1(g) implies that α 2 = 1, so 4 = p 2 1 = S 2 = 1 + p 2, so p 2 = 3. This implies that 9 = p 2 2 = S 1 = 1 + p p α1 1, but there is no integer α 1 which makes this equality true. Hence there are no values of n where σ = γ 2, if = 2. emma 3. If = 3 and σ = γ 2, then n = 1782 = Proof: The general method of proof will be determining which of the prime factors p 1, p 2, and p 3 divide evenly into each of S 1, S 2, and S 3 and whether p 2 1, p 2 2, or p 2 3 divides evenly into any of them. There are only a finite number of ways for this to happen, since there are only six prime factors and three values of S i (p 2 1p 2 2p 2 3 = S 1 S 2 S 3 ) one possibility is that S 1 = p 2, S 2 = p 2 3, and S 3 = p 2 1p 2. Then some bounds will be determined for α i from these equations; it turns out that certain exponents cannot be too large without a contradiction occuring in the example given, it is shown that α 2 is 2 or 4, because the assumption that α 2 6 lead to contradictions. Finally, the information about some α i will be used to find out further information about the values of the other S i s, and the primes p i ; most of the time, this new information will lead to a quick contradiction. In the example given, it is shown if α 2 = 4, then p 2 = 3, α 1 = 1, S 2 = 121, p 3 = 11, and n = 1782 (which are deduced in this order); if α 2 = 2, then a contradiction results (p 3 must be 3/5 or 0, neither of which is a prime number). It is likely that using this technique, the conjecture may be resolved for values of n with > 3. Now suppose σ = γ 2. Note that p 1 S 3, because otherwise, S 3 = p 2 or S 3 = p 2 2, which do not happen, according to emma 1(h) and (i). Also because of emma 1(h) and (i), S 3 p 1, p 2 1, so S 3 = p 2 1p 2 2 (Case I), S 3 = p 1 p 2 2 (Case II), S 3 = p 1 p 2 (Case III), or S 3 = p 2 1p 2 (Case IV). Furthermore, α 3 = 1, by emma 1(g). We eliminate Case I first; if S 3 = p 2 1p 2 2, p 3 = S 3 1 = p 2 1p = (p 1 p 2 1)(p 1 p 2 + 1), which is a non-trivial factorization of p 3, as p 1 p > 1. Now for Case II. Since p 2 1 S 3, p 1 S 2 ; otherwise p 1 S 1, contradicting emma 1(c). Since S 2 p 1, p 3 S 2 as well, because of emma 1(h). If p 2 3 S 2, then, in order for σ to equal γ 2, we must have S 1 = 1, which contradicts emma 1(h). Thus p 3 S 1, and we have S 1 = p 3 and S 2 = p 1 p 3 ; in particular, p 3 = S 1 = α1 = 2 α1+1 1, and so 2p 2 2 = p 1 p 2 2 = S 3 = 1 + p 3 = 2 α1+1, but then p 2 divides evenly into the left-hand side and not into the right-hand side. This takes care of Case II. 4

5 Now consider Case III (S 3 = p 2 1p 2, noting that p 3 = 4p 2 1). Since S 1 S 2 S 3 = p 2 1p 2 2p 2 3 and p 2 S 2, p 2 S 1. Similarly, p 3 S 2. Thus we either have S 2 = p 2 3 and S 1 = p 2 (Case IIIA) or S 2 = p 3 and S 1 = p 2 p 3 (Case IIIB). For Case IIIA: S 2 = p 2 3 = (4p 2 1) 2, and α 2 is even. If α 2 4, then (4p 2 1) 2 = p 2 3 = S 2 p 4 2 [> 0], so 4p 2 1 p 2 2, which implies that p 2 = 3, and consequently p 3 = S 3 1 = 4p 2 1 = 11. We also have α 2 = 4, because if α 2 6, Since 121 = 11 2 = (4p 2 1) 2 = S 2 > p 6 2 = 3 6 = α1+1 1 = 1 + p p α1 1 = S 1 = p 2 = 3, α 1 = 1. Thus n = = If α 2 = 2, then (4p 2 1) 2 = p 2 3 = S 2 = 1 + p 2 + p 2 2, which implies that p 2 = 3/5 or 0, neither of which is prime. This settles Case IIIA. For Case IIIB: S 2 = p 3 has no p 1 factors, so by emma 1(e), α 2 is even and at least two. Hence 1 + p 2 + p 2 2 S 2 = p 3 = 4p 2 1, which implies that p 2 2 3p , which only occurs if 1 p 2 2. Since p 2 > p 1 = 2, this is a contradiction. This takes care of Case III. Now we turn to Case IV, where S 3 = p 1 p 2 and hence p 3 = 2p 2 1. Because there needs to be another p 1 factor and another p 2 factor in S 1 S 2 S 3, p 1 S 2 and p 2 S 1. Since p 3 S 3, we must have S 2 = p 1 (which cannot occur, by emma 1(h)), S 2 = p 1 p 3 (Case IVA), or S 2 = p 1 p 2 3 (Case IVB). We consider Case IVA: If α 2 5, then 2(2p 2 1) = p 1 p 3 = S 2 > 1 + p p 5 2 > p , or 0 > p 2 2 4p = (p 2 2) 2. Thus, α 2 5, and so by emma 1(f), α 2 = 1, and 1 + p 2 = S 2 = 2(2p 2 1) = 4p 2 2, which implies that p 2 = 1. This is not possible, either. Finally, consider Case IVB. If α 2 5, then 8p 2 2 8p = 2(2p 2 1) 2 = p 1 p 2 3 = S 2 > 1 + p 2 + p 5 2 > p 3 2 = 2 + 2p 3 2, which implies that 8p 2 2 8p 2 > 2p 3 2, or Thus α 2 = 1 by emma 1(f), and so 0 > 2p 2 8p = 2(p 2 2) p 2 = S 2 = p 1 p 2 3 = 2(p 2 1) 2 = 8p 2 2 8p 2 + 2, or 0 = 8p 2 2 9p = (8p 2 1)(p 2 1), which implies that p 2 is 1 or 1/8; but neither of these numbers is prime. This settles all possible cases, and proves emma 3. 5