{VERSION 6 0 "IBM INTEL NT" "6.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 0 128 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 128 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 128 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Map le Output" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 128 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 128 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 226 "Searching for special dou ble solutions to 2^x + 2^y = p^z - p^w for prime p, used in paper \"O n the generailzed Pillai equation $\\pm a^x \\pm b^y = c$\" where we h ave the embarrassing unfinished case in Theorem 6 of p = 1 mod 16" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "March 18, 2004" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 134 "A solution which is problematic for our results satisfy p = 1 mod 16. We can use three different approaches to eliminate values of p." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 378 "The firs t is simple congruence conditions: if r is max such that 2^r | p-1, t hen if r = 4 can eliminate all but p mod 5 = 2, then if r=4 can elimin ate all except p mod 16 = -2, -6, or 8. If r = 5, 8, 9, 12, or 13 etc . can eliminate all but p mod 5 = 2, if r = 6, 7, 10, 11, etc, can eli minate all but p mod 5 = 3. These conditions seem to eliminate about \+ 2/3 of all primes. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 551 "Second, we can rewrite 2^x + 2^y = p^z - p^w as 2^x( \+ 2^(y-x) + 1 ) = p^z (p^(w-z) - 1 ) and we claim that y-x is odd and th at z-w must be 2 mod 4. In particular, z-w is even so p and p-1 and p +1 divide both sides. Let q be the odd part of p-1, and let r be th e odd part of p+1. Define minusorder(b,a) to mean the least power t s uch that b^t + 1 = 0 mod a, and let it be false if there is no such or der. Then we want minusorder(2,p), minusorder(2,q) and minusorder(2, r) to all be odd. We search for any prime p satisfying all these cond itions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 708 "Thirdly, we use Theorem 2 equation 13 of Scott's 1993 paper which gives a bound in terms of class numbers. Let N = p + 2^(r+1) ; let P = product of all primes dividing N to an odd power, and Q be the prod uct of all primes dividing N to an even power. Define L in terms of t he lcm of the Q factor as in Theorem 2. If r mod 2 = 0 then let h be \+ the class number of Q( sqrt(-P) ), if r mod 2 = 1 then h is class numb er for 2*P instead. Then consider the set of divisors of J = L*h. If any divisor of L is congruent to 3 or 7 mod 16 and also not 1 mod 3, \+ then call such a divisor k; we check if p^k - p - 2^(r+1) is a power of 2. Actually we only check if it is congruent to zero mod a large \+ power of 2. " }}{PARA 0 "" 0 "" {TEXT -1 96 "\nWe could also use boot strapping, but this turns out to take longer and be less efficient. \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "restart; with(numtheory):" }}{PARA 7 "" 1 "" {TEXT -1 69 "Warning, the protected name order has been redefined and unprot ected\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Here are the procedure s to use the divisors p, p-1, p+1 approach." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "We find the odd part of a number." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "oddpart := pr oc(n)\nlocal a, i;\na:= n;\nwhile a mod 2 = 0 do a := a/2; od;\na;\nen d;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(oddpartGf*6#%\"nG6$%\"aG%\"iG 6\"F+C%>8$9$?(F+\"\"\"F1F+/-%$modG6$F.\"\"#\"\"!>F.,$*&#F1F6F1F.F1F1F. F+F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 166 "We use the \"order\" t hat gives -1 if any, that is, minusord = t means t is the least numbe r such that p divides 2^t+1. \nIf there is no such t, we set it to \" fail\". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "minusorder := proc(a,b)\nlocal i,ans,k,t;\ni:= order(a,b);\nans := false;\nwhile i \+ mod 2 = 0 and ans=false do \ni := i/2;\nif b^i +1 mod a = 0 then ans : = i; fi;\nod;\nans;\nend;\n " }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%+min usorderGf*6$%\"aG%\"bG6&%\"iG%$ansG%\"kG%\"tG6\"F.C&>8$-_%*numtheoryG% &orderG6$9$9%>8%%&falseG?(F.\"\"\"F=F.3/-%$modG6$F1\"\"#\"\"!/F:F;C$>F 1,$*&#F=FCF=F1F=F=@$/-FA6$,&)F8F1F=F=F=F7FD>F:F1F:F.F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Now for the least effective but easiest congruence approach. \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 134 "We d efine congruence conditions that help eliminate possibilities. The co nditions are determined by how many factors of 2 divide p-1." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "twoexponent:= proc(n)\nlocal i,k;\nk:= n; i:= 0;\nwhile k mod 2 = 0 do\ni:= i+1; k := k/2;\nod;\ni ;\nend;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,twoexponentGf*6#%\"nG6$% \"iG%\"kG6\"F+C&>8%9$>8$\"\"!?(F+\"\"\"F4F+/-%$modG6$F.\"\"#F2C$>F1,&F 1F4F4F4>F.,$*&#F4F9F4F.F4F4F1F+F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "twoexponent(64);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# \"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 464 "congruenceconditi on := proc(p)\nlocal r, ans, p5,p17;\nr := twoexponent(p-1);\np5 := p \+ mod 5; \nans:= true;\nif r<4 then print(\"problem\");fi; #this should never happen since p mod 16 = 1\nif r = 4 then if p5 <> 2 then ans:= \+ false \nelse \np17 := p mod 17;\nif (p17 <> 15 and p17 <> 11 and p17 < > 8) then ans:= false \nfi;fi;fi;\nif r = 5 or r=8 or r=9 then if p5<> 2 then ans:= false\nelse\nif r=6 or r=7 or r=10 or r=11 then if p5<>3 \+ then ans:= false;\nfi;fi;fi;fi;\nans;\nend; \n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%4congruenceconditionGf*6#%\"pG6&%\"rG%$ansG%#p5G%$p17 G6\"F-C)>8$-%,twoexponentG6#,&9$\"\"\"F6!\"\">8&-%$modG6$F5\"\"&>8%%%t rueG@$2F0\"\"%-%&printG6#Q(problemF-@$/F0FC@%0F9\"\"#>F?%&falseGC$>8'- F;6$F5\"#<@$330FQ\"#:0FQ\"#60FQ\"\")>F?FN@$55/F0F=/F0Fgn/F0\"\"*@%FK>F ?FN@$555/F0\"\"'/F0\"\"(/F0\"#5/F0Fen@$0F9\"\"$>F?FNF?F-F-F-" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "congruencecondition(9137);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 207 "counttrue:= 0; countfalse:=0;\nfor p from 17 to 10 ^7 by 16 do\nif isprime(p) then\nif congruencecondition(p) = true then counttrue:= counttrue+1; else countfalse:= countfalse+1; fi;fi;od;\nc ounttrue; countfalse;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*counttrueG \"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+countfalseG\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"&#*\\#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# \"&vz&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 161 "This took about 15 min utes. \nThis shows that about two thirds of the primes are taken car e of by the congruence condition which seems to be the fastest also. \+ " }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 93 "Now we use a class number condition test: first we find the L factor of t he square part of N." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 227 "ge tPQ := proc(n)\nlocal P, Q, f2, k, b, a, i;\nf2 := op(2, ifactors(n)); \nk := nops(f2);\nP := 1; Q:= 1;\nfor i from 1 to k do\nb := op([i,1], f2);\na := op([i,2], f2);\nif a mod 2 = 1 then P := P*b; else Q := Q* b; fi;\nod;\n[P,Q];\nend;\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&getP QGf*6#%\"nG6)%\"PG%\"QG%#f2G%\"kG%\"bG%\"aG%\"iG6\"F0C(>8&-%#opG6$\"\" #-%)ifactorsG6#9$>8'-%%nopsG6#F3>8$\"\"\">8%FC?(8*FCFCF=%%trueGC%>8(-F 56$7$FGFCF3>8)-F56$7$FGF7F3@%/-%$modG6$FPF7FC>FB*&FBFCFKFC>FE*&FEFCFKF C7$FBFEF0F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "getPQ(2^30 *3^20*5^21);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$\"\"&\"\"'" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 180 "getL := proc(P,Q)\nlocal L, Qfac, i, q;\nL := 1;\nif Q > 1 then\nQfac := factorset(Q);\nfor i fro m 1 to nops(Qfac) do \nq := op(i,Qfac);\nL := lcm(L, q - legendre(-P, \+ q));\nod;\nfi;\nL;\nend;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%getLGf* 6$%\"PG%\"QG6&%\"LG%%QfacG%\"iG%\"qG6\"F.C%>8$\"\"\"@$2F29%C$>8%-_%*nu mtheoryG%*factorsetG6#F5?(8&F2F2-%%nopsG6#F8%%trueGC$>8'-%#opG6$F?F8>F 1-%$lcmG6$F1,&FFF2-_F;%)legendreG6$,$9$!\"\"FFFUF1F.F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "getL(3*5*11, 5*13);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#g" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 411 "imagquadclassnum := proc(N)\nlocal i, h, a, b, b1, be, d;\n#we a ssume here that N is squarefree\nif N < 4 then h := 1; else\nif N mod \+ 4 = 3 then \nd := N; b := 1;\nelse \nd := 4*N; b := 0;\nfi;\nh := 0;\n for b1 from b to isqrt( iquo(d,3) ) by 2 do\nbe := iquo( b1^2 + d, 4); \nfor a from max(1, b1) to isqrt(be) do\nif be mod a = 0 then h := h+1 ; \nif b1 <>0 and a<>b1 and a^2 <> be then h := h+1; fi;\nfi;\nod;\nod ;\nfi;\nh;\nend;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%1imagquadclassnu mGf*6#%\"NG6)%\"iG%\"hG%\"aG%\"bG%#b1G%#beG%\"dG6\"F0C$@%29$\"\"%>8%\" \"\"C%@%/-%$modG6$F4F5\"\"$C$>8*F4>8'F8C$>FB,$*&F5F8F4F8F8>FD\"\"!>F7F J?(8(FD\"\"#-%&isqrtG6#-%%iquoG6$FBF?%%trueGC$>8)-FS6$,&*$)FMFNF8F8FBF 8F5?(8&-%$maxG6$F8FMF8-FP6#FXFU@$/-F=6$FXFinFJC$>F7,&F7F8F8F8@$330FMFJ 0FinFM0*$)FinFNF8FX>F7FeoF7F0F0F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "imagquadclassnum(1234567);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"$'\\" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "This test is based on Theorem 2 \+ of Scott's 1993 paper. " }}{PARA 0 "" 0 "" {TEXT -1 244 "We use a bou nd of 2^(40+r), below which we calculate the full p^k-p-2^(r+1) and se e if it is a perfect power of 2, but above this bound we just view thi s modulo 2^8 to see if it is a power of 2. If it is 0 mod 2^8, we als o check it mod 2^15. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 627 " classnumtest := proc(p)\nlocal r, n, PQ, P, Q, L, J, k, divlist, w, i; \nr := twoexponent(p-1);\nn := p + 2^(r+1);\nPQ := getPQ(n);\nP := op( 1, PQ); Q := op(2, PQ);\nL := getL(P,Q);\nif r mod 2 = 0 then J := L*i magquadclassnum(P); \nelse J := L*imagquadclassnum(2*P); fi;\ndivlist := divisors(J);\nfor i from 1 to nops(divlist) do\nk := divlist[i];\n if (k mod 16 = 3 or k mod 16 = 7) and k mod 3 <> 1 then \nif k < evalf ((40+r)*log(2)/log(p)) then \nw := p^k - p - 2^(r+1);\nif oddpart(w) = 1 then w := 0; fi;\nelse\nw := (p&^k - p - 2&^(r+1)) mod 2^8;\nfi;\ni f w = 0 then print(p, k, r, n, J, (p&^k - p - 2&^(r+1)) mod 2^15); fi; \nfi;\nod;\nend; \n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%-classnumtes tGf*6#%\"pG6-%\"rG%\"nG%#PQG%\"PG%\"QG%\"LG%\"JG%\"kG%(divlistG%\"wG% \"iG6\"F4C+>8$-%,twoexponentG6#,&9$\"\"\"F=!\"\">8%,&F8&-%&getPQG6#F@>8'-%#opG6$F=FF>8(-FM6$FCFF>8)-%%getLG6$FKFP@%/-% $modG6$F7FC\"\"!>8**&FTF=-%1imagquadclassnumG6#FKF=>Fin*&FTF=-F\\o6#,$ *&FCF=FKF=F=F=>8,-_%*numtheoryG%)divisorsG6#Fin?(8.F=F=-%%nopsG6#Feo%% trueGC$>8+&Feo6#F\\p@$35/-Fen6$Fcp\"#;\"\"$/Fjp\"\"(0-Fen6$FcpF]qF=C$@ %2Fcp-%&evalfG6#*(,&\"#SF=F7F=F=-%$logG6#FCF=-F]r6#FC$>8-,()FFBF>@$/-%(oddpartG6#FcrF=>FcrFgn>Fcr-Fen6$,(-%#&^G6$F-Fa s6$FCFDF>\"$c#@$/FcrFgn-%&printG6(F " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 342 "for p from 17 to 10^8 by 16 do\nif isprime (p) then\nif congruencecondition(p) = true then\nif (2&^oddpart(order( 2,p)) + 1) mod p = 0 then \nq:= oddpart(p-1);\nif q>1 then\nif (2&^odd part(order(2,q)) + 1) mod q = 0 then \nr := oddpart(p+1);\nif r>1 then \nif (2&^oddpart(order(2,r))+ 1) mod r = 0 then \nclassnumtest(p);\nfi ;\nfi;\nfi;\nfi;\nfi;\nfi;\nfi; \nod;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(xPC#\"\"$\"\"'\"(0RC#\"$#z\"&'\\5" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(xPC#\"#**\"\"'\"(0RC#\"$#z\"&G*G" }}{PARA 11 "" 1 " " {XPPMATH 20 "6(\"(L7N$\"\"$\"\"'\"(h8N$\"%G<\"&_*H" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(\"\"\"%\"(\\KY$\"%G9\"%;G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"((*GJ&\"\"$\"\"(\"(`JJ&\"%KA\"&)3D" }}{PARA 11 " " 1 "" {XPPMATH 20 "6(\"((*GJ&\"$z#\"\"(\"(`JJ&\"%KA\"&%Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(\"\"\"%\"(4#oj\"%G9\"&GD#" }} {PARA 11 "" 1 "" {XPPMATH 20 "6(\"(t1=(\"\"$\"\"(\"(H4=(\"%[O\"&)3D" } }{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(dY$z\"\"$\"\"'\"(&yMz\"%%e\"\"&g<# " }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(dY$z\"#**\"\"'\"(&yMz\"%%e\"\"& 3Q#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"((4c*)\"\"$\"\"'\"(Di&*)\"%'H \"\"$c#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(L/(*)\"\"$\"\"'\"(h0(*) \"%G6\"&G*G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\"(d!H**\"#R\"\"&\"(@\" H**\"%WP\"&S-\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")Pb25\"\"$\"\"%\" )pb25\"%oJ\"&kS#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")Pb25\"#**\"\"% \")pb25\"%oJ\"&+c#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")<;h7\"\"$\"\" &\")\"o6E\"\"%g@\"&[w#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")d9'H\"\"# ()\"\"%\")*[hH\"\"%)3#\"&3u\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")(H ,<#\"#B\"\"%\")H8q@\"%7L\"&wX#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")( HL@#\"$F$\"\"%\")HL8A\"%;E\"&?z\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6( \")PR7G\"\"$\"\"&\"),S7G\"%)o#\"%3Y" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 (\")x`iH\"\"$\"\"&\")TaiH\"%CU\"%;#*" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6(\")L;5I\"#N\"\"(\")*)=5I\"%?R\"&s'G" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")PR&3$\"#R\"\"%\")pR&3$\"%cS\"&7h#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")P:]K\"\"$\"\"&\"),;]K\"%g@\"&_2\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")%\"\"&\")T#yM$\"%_L\"&[w#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")dUiL\"\"$\"\"'\")&QCO$\"%G<\"&O2#" }} {PARA 11 "" 1 "" {XPPMATH 20 "6(\")Pz=O\"\"$\"\"&\"),!)=O\"&k;\"\"&_2 \"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")Pz=O\"$V#\"\"&\"),!)=O\"&k;\" \"&OV\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")d=hO\"\"$\"\"%\")*)=hO\" %/f\"$c#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")x4SP\"\"$\"\"%\")45SP\" %ga\"&?z\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")x4SP\"#N\"\"%\")45SP \"%ga\"&K%=" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")x4SP\"$&>\"\"%\")45S P\"%ga\"&#*4#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")xtxR\"$N\"\"\"&\") TuxR\"%?V\"&kS#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")L)3+%\"\"$\"\"( \")*34+%\"%sO\"&OV\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")L)3+%\"#^\" \"(\")*34+%\"%sO\"%'4%" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")d%**4%\" \"$\"\"&\")@&**4%\"%Sc\"&%Q;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\") " 0 "" {MPLTEXT 1 0 347 "for p from 73152817 to 10^8 by 16 do\nif isprime(p) then\nif cong ruencecondition(p) = true then\nif (2&^oddpart(order(2,p)) + 1) mod p \+ = 0 then \nq:= oddpart(p-1);\nif q>1 then\nif (2&^oddpart(order(2,q)) \+ + 1) mod q = 0 then \nr := oddpart(p+1);\nif r>1 then\nif (2&^oddpart( order(2,r))+ 1) mod r = 0 then \nclassnumtest(p);\nfi;\nfi;\nfi;\nfi; \nfi;\nfi;\nfi; \nod;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")\"\"\"'\")@z\"*\"\"$ \"\"(\")H,?\"*\"&O6\"\"&%=H" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")t)*> \"*\"#()\"\"(\")H,?\"*\"&O6\"\"&![?" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 (\")d%RD*\"\"$\"\"'\")&eRD*\"%+[\"&3u\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")<3!R*\"\"$\"\"%\")\\3!R*\"%Gx\"&cI\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6(\")<3!R*\"$$[\"\"%\")\\3!R*\"%Gx\"$c#" }}}{EXCHG {PARA 257 "" 1 "" {TEXT -1 100 "We see the only case we need to check is 791 77793. We check the value of p^k-p-2^(r+1) modulo 2^16:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "p:= 79177793; k:= 3; r := 6; x := p ^k - p - 2^(r+1); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pG\")$zx\"z " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG\"\"$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG\"\"'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG\" 9O$=BC]Ah:`P'\\" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "x mod 2^ 15; x mod 2^16;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#\"&oF$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "32" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }