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Internat. J. Math. and Math. Sci. Vol. i0 No. 2 (1987) 409-412 409 EVEN PERFECT NUMBERS AND THEIR EULER’S FUNCTION SYED ASADULLA Department of Mathematics and Computin Sciences St. Francis Xavier University Antigonish, Noga Scotia. (Received February 21, 1985, and ABSTRACT. B2G 1C0. Canada. in revised form, June 23, 1986) The purpose of this article is to prove some results on even perfect numbers and on their Euler’s function. The results obtained are all straightfor- ward deductions from well-known elementary number theory. KEY WORDS AND PHRASES. Perfect number; triangular number- Euler’s function; number of divisors function. 1980 MATHEMATICS SUBJECT CLASSIFICATION CODE. INTRODUCTION. A positive integer 1. IOAdO is called a perfect number if it is equal to the sum of its positive divisors excluding itself. The n th triangular number is the sum of the first n-positive integers I k F . n k=l Euler’s function to @(n) is the number of positive integers less than or equal and relatively prime to n T(n). n(n+l) n. The number of divisors function is the number of positive divisors of d(n) MAIN RESULTS. The proof of the following Theorem can be found in many elementary number theory books; see, for example, [l:p. 98]. THEOREM 1. If n is an even perfect number, there exists a prime that THEOREM 2. Pk If T(pl) is any even perfect number, where is the first prime in the sequence then 2P-1 such p-1(2p-1). 2 n T(Pk) {P2, P3 Pj Pl is prime, and if } where pj 2Pj_1+1, is the next even perfect number. It follows from Theorem 1 that an even perfect number is of the form can be written as T(pI), 2n-I is prime. Now, =2n-1. Let Pi be any composite term of the sequence PROOF. 2n-l(2n-l), where pl ..... Pi It can be shown that Pj 2n-l(2n-l) where 2Pj_1+1. n+i-1 -1, 2 Now, it follows from Theorem using the facts that T(Pi) Pi 2n-1’ n+i-1 2n+i-2(2 and -I) is 410 S. ASADULLA Let not an even perfect number. As before, }. pj 2m-1(2m-l), is of the form 2n+k-I Pk T(Pk) 2n+k-2(2 thus T(Pk) is an even n+k-I Observe that -1. 2m-1 where (P2, be the first prime in the sequence Pk is prime and P3, -1) perfect number by Theorem 1. EXAMPLE. TC3 C3)Cd) T(31) THEOREm3. C31)(32) If n= then, 2 (m+1)/2 2k, + (2k-1) 2 (m-1)/2 (127)(128) + 3 1 n + + + 3 + 2 [2 + 4 + 2o(1 o + 2 + k(2k+l) 2k-(k+l) + [2 (m+l)/2 -1] 3. Now, consider =[ (2k)(2k+l)]2 (2k-1) + + C2k) ] + k) + k) 8(I + 2 + S[ k(k+l)]Z k(2k 1). follows that 1+ 3 +...+ [2 {m+1)12 it 8128 2 {m-1)/2. k where k-C2k+l) k(2k+l) k-(2k+l) kZ(2k+l) z 2 28. [1+2+3+...+(2k-1) + (2k)] 2 + (2k) which implies that k T(127) 496 2m-l(2m-l), Observe that Since C7)C 8) TO7 6, 2m-1(2m-l)-- 1] 3 The following Corollary I, follows from Theorem 3. COROLLARY 1. If n n EXAMPLE. is an even perfect number + 3 1 + 3 496 + + C2 (p+I)/2 + 7; p + 5 2P-1(2P-1), 1] . then 5. The proof of the followingTheorem 4 can also be found in many elementary 1: number theory books; see, for example THEOREM 4. If n Pt 51 63. p. Pk% P2a 1. 1 P P’ P are distinct primes and Pk - a a, a2 are positive integers. k As a consequence of Theorem 4, one can easily obtain Theorem 5, Corollary 2, and Corollary 3 2P-1(2p #(.) P-(2 THEOREM 5. n COROLLARY 2. EXAtLE. It n THEORE 6. If n If then PROOF. @(nt pi-1 @C 2 n, n..., pi. 4032 8128 n @(n) n 4p-1. @(n) n 2p-2. 46 is an even perfect number, then are k-distinct even perfect numbers, n k nk)= .k-1 @(nk). @(nt) @(n) nk) p.-1 (2p -1) 2 + @E2 PI+P- + @(2 P- ,h, is an even perfect number, then @(nl n2 PI+P- + ). @(26 @(127) @{S128) COROLLARY 3. is an even perfect number if and only if 1) + Pk-k Pk-k Pk-1 2p- -1) 2 P Pl -I) (2 (2 Pl @(2 Pk -I)] Pk @(2 -1) (2 -1) P- -1) @(2 pk (2 -1)] -I) n EVEN PERFECT NUMBERS AND THEIR EULER’S FUNCTION + Px+P+ -k-1 Pk =2 p p -2) (2 (2 (2Pk -2) -2) p-I p-I pi-1 2 {2 -1) k-1 2 p-I {2 2 411 Pk-1 -1) 2 2Pk-1 -1) k-1 @(nl) @On2) 9. @(nk). The following Theorem 7 is proved in many books on elementary number theory; see, for example, [1: p. If n 9]. k THEORE 7. a R i=1 k are distinct primes and k i a R (I + d{n) then Pi i=1 Pi’ i=l d(n) are positive integers, and k i=1 i. where ai), is the number of divisors function. k THORE 8. If a R n i=1 2p-I(2p i and Pi d{n} is an even perfect number I), then i} p ii} a 2 uj o. a 2 iil) k. j I k iv) for exactly one -I, I> where O, k ff i=l (1 + aj). J k, Thus (1 + and (2p -1) p k {1 # J, / I > 2p-l, t) J X O; k {1 + a p-1 2 (2p -1), i) which tmplie that 0 gj 1 (l+ai), and i=1 / and j which say such that that is, U 2 i p-I i k, i X for some X and exactly one (2 -1) aj) i=l X k p-1 2 (2p -1), (l+ai) (2p -1) divides exactly one of the factors implies that < J J. I k, 1 From Theorem 5, one obtains d(n) PROOF. 1 such that 1. p Ui i=1 {2p -I} -I which is k Observe that p-1 k-1 or p for exactly one (2p 1) 1 , k-1 i=1 i k, which is such that 1 for exactly one J since i > (t). Now, {1 + aj) {2p -1) J k and j 0 implies that such that 0 1 for i k J and O. and O, aj Thus, 2J which proves 412 S. Finally, a =2 1, 1 ACKNOWI. + a 2 i k, 1 i <_ J, i ASADULLA k, i > i O, J, i > 0 implies that which proves (tit). The author wishes to tl-tk the referee for his many helpful sugge s t ions. I. SHOCKL, J.E., York, N.Y. REFERENCES Introduction to Number Theory, Holt, Rinehart and Winston, New 1967.