Mrgolius 2 In the rticulr cse of Ploue's constnt, we tke = 2= 5+i= 5, nd = ;1, then ; C = tn;1 1 2 = ln(2= 5+i= 5) ln(;1) More generlly, we would hve

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1 Ploue's Constnt is Trnscendentl Brr H. Mrgolius Clevelnd Stte University Clevelnd, Ohio Astrct Ploue's constnt tn;1 ( 1 2) is trnscendentl. We rove this nd more generl result using sequences of rimitive ythgoren triles. Ploue's constnt [6], [2], C = tn;1 ; is trnscendentl. In this note, we rove this nd the more generl result tht Theorem 1 tn ;1 (x) is trnscendentl when x is rtionl nd x 6= 0or 1. The result is consequence of the Gelfond-Schneider Theorem which ws roved indeendently y A.O. Gelfond nd Th. Schneider in 1934 solving Hilert's seventh rolem from the list of twenty-three outstnding rolems tht he nnounced in The seventh rolem delt with the irrtionlity nd trnscendnce of certin numers. One form of the Gelfond-Schneider Theorem is Theorem 2 (Gelfond-Schneider) If nd re non-zero lgeric numers, nd if 6= 1 then (log())=(log()) is either rtionl or trnscendentl. This is theorem 10.2,. 135 from Niven, [5]. References re rovided in Niven, [5],. 149.

2 Mrgolius 2 In the rticulr cse of Ploue's constnt, we tke = 2= 5+i= 5, nd = ;1, then ; C = tn;1 1 2 = ln(2= 5+i= 5) ln(;1) More generlly, we would hve = = i= 2 + 2, where nd re reltively rime integers, nd tn ;1 ; = ln(= i= ) ln(;1) nd re lgeric (they re solutions of olynomil equtions with integer coecients), so to show thttn ;1 (=)= is trnscendentl, it remins to show tht tn ;1 (=)= is irrtionl. This result lso is given in Niven. By corollry 3.12, [5],. 41, if = 2r for some rtionl numer r, then tn is rtionl only if tn = 0 1. Niven sttes this result s corollry to theorem due to D.H. Lehmer. We rove this result y other mens. Theorem 3 tn ;1 (x) is irrtionl when x is rtionl nd x 6= 0or 1. Proof Suose not, then for some reltively rime ositive integers r nd s, tn ;1 (x) = r, nd so s tn;1 (x) = r. If this is so, then s tn 2s tn ;1 =tn(2r) =0 We show tht no ositive integer s with this roerty exists. Let x = where nd re reltively rime integers. Recll the identity tn(u + v) = tn(u)+tn(v) 1 ; tn(u)tn(v) (1) We dene the sequence n =tn 2n tn ;1 which, from eqution (1), stises the recursion n = k + n;k 1 ; k n;k

3 Mrgolius 3 for k =1 n;1. Wecnnd 1 y lying eqution (1) with u = v = tn ;1 ; tn 2tn ;1 = = 1 = = ; 2tn ; ; tn ;1 ; 1 ; tn tn ;1 2 2= 2 1 ; ; 2 2 ; 2 Let 1 =2, ndq 1 = 2 ; 2 if 2 ; 2 is odd, nd let 1 =, ndq 1 =( 2 ; 2 )=2 if 2 ; 2 is even. Then 1 nd q 1 re reltively rime. (If this were not so, then there would e rime numer which divides oth nd hence either or, nd which divides either ; or +. A rime which divides () nd ; must lso divide (), nd similrly rime which divides () nd + must divide (). But this contrdicts our ssumtion tht nd re reltively rime.) The solute vlues of the numers ( 1 q q2 1 ) form rimitive ythgoren trile, tht is, the numers ( 1 q q2 1 ) re integers which hve no common fctor nd their solute vlues reresent sides of ythgoren tringle. Ernest Eckert [1] nd others [4],[8] hve shown tht the set of rimitive ythgoren triles is grou under the oertion clled ddition dened s follows, Theorem 4 (Eckert) The set P of rimitive ythgoren triles is grou under the oertion clled ddition, dened y ( c)+(a B C) = 8 < (A ; B A + B cc) when A ; B > 0 (A + B B ; A cc) when A ; B 0 The identity element in P is (1 0 1), nd the inverse of ( c) is ( c). >From this it follows esily tht the set of ordered triles of integer comonents the rst two of which my e either ositive or negtive ut whose solute vlues form rimitive ythgoren triles form grou under ddition dened y ( c)+(a B C) =(A + B B ; A cc) with the identity element (0 1 1) nd inverse of ( c) equl to (; c). Let n = n qn where n nd q n re reltively rime. Then n+1 = 1 + n q1 qn 1 ; 1 q1 n qn

4 Mrgolius 4 Alying Eckert's theorem, = 1q n + n q 1 q 1 q n ; 1 n nd n+1 = 1 q n + n q 1 q n+1 = q 1 q n ; 1 n re reltively rime, q 2 + n+1 q2 n+1 =(q ) (n+1)=2 q is n integer nd (j n+1 j jq n+1 j 2 + n+1 q2 n+1 ) is rimitive ythgoren trile. The numer n cnnot e equl to zero for ny numer n, nd hence, no ositive integer s exists such tht so tn 2s tn ;1 tn ;1 (x) =tn(2r) =0 is irrtionl when x is rtionl nd x 6= 0 or 1. This comletes the roof of theorem 3, nd this result together with the Gelfond-Schneider theorem roves theorem 1. As corollry, we hve tht Corollry 1 The sequences n nd q n given y the formuls n = ( ) n sin q n = ( ) n cos 2n tn ;1 2n tn ;1 (for 6= 0, nd 6=, nd oth integers) re integer sequences, nd 2 n + q 2 n = ( ) 2n for ll n =1 2 3, thus (j n j jq n j ( ) n ) form n innite sequence of ythgoren triles. In rticulr, we hve the sequences of ordered triles (sequences A066647, A nd A in EIS[7])

5 Mrgolius 5 (4 3 5) (24 ;7 5 2 ) (44 ; ) (;336 ; ) (;3116 ; ) (; ) ( ) ( ) ( ; ) ( ; ) (; ; ) (; ) (; ), ; ; for the sequence generted y tn 2n tn ;1 1 2,nd ( ) (120 ; ) (;828 ; ) (;28560 ; ) (; ) ( ) ( ; ) ( ; ) (; ; ) (; ) ( ) ; ; for the sequence generted ytn 2n tn ; (sequences A067358, nd A in EIS[7]). Eckert[1],. 26, lso showed tht there is one nd only one rimitiveythgoren trile for ech hyotenuse of the form n where 1 mod 4 is rime nd we do not distinguish etween tringles ( c) nd ( c), so ech of these rimitive ythgoren triles is unique. References [1] Ernest J. Eckert, The Grou of rimitive Pythgoren tringles, Mthemtics Mgzine 57 (1984) [2] Ploue's constnt t the site Fvorite Mthemticl Constnt ofsteve Finch, htt//uillc.inri.fr/lgo/solve/constnt/l/l.html, [3] Rhel Finkelstein, rooser nd J. C. Lgris, solver, Prolem 6053, Amer. Mth. Monthly, 84 (1977) 493. [4] J. Mrini, The grou of the Pythgoren numers, Amer. Mth. Monthly, 69 (1962) [5] Ivn Niven, Irrtionl Numers, The Crus Mthemticl Monogrhs, Numer 11, second rinting, 1963, The Mthemticl Assocition of Americ, John Wiley & Sons, New York, NY. [6] S. Ploue, The comuttion of certin numers using ruler nd comss, J. Integer Sequences 1 (1998) MR 2000c11211.

6 Mrgolius 6 [7] N. J. A. Slone, The on-line encycloedi of integer sequences, ulished electroniclly t htt// njs/sequences/, [8] Olg Tussky, Amer. Mth. Monthly, 77 (1970)