Herewith the solution to Oyler's X-number puzzle posted 24 March 4128624 2012599 7138444 1169525 7595606 rgdz Pete Kogel I solved the clues in approximately the order 7, 8, 2, 1, 3, 5, 6, 4. All solutions to the equations can be reduced to three variables as in the first clue, unless a factor can be divided out. Then, the resulting solution is of the same form. The actual answer is: 4128624 2012599 7138444 1169525 7595606 Bryce Herdt (a,b,c,d) cd=ab a^2+b^2=c^2 c=xy d=zw a=xz b=yw (y,z)=1 (x,w)=1 x^2z^2+y^2w^2=x^2y^2 y^2|x^2z^2 => y^2|x^2 x^2|y^2w^2 => x^2|y^2 => x=y z^2+w^2=x^2 a=xz, b=xw, c=x^2, d=zw 7.(k,C,K,A) Each of them less than 100 and more than 10. x<10, z^2+w^2=x^2 => x=5, {z,w}={3,4} Therefore K=25 and A=12. {k,C}={15,20}. But k cannot be 20 as M cannot start with 0. Thus k=15 and C=20. .12.... 20..... ....... .1...25 .5..... 8.(m',A(K-A),(K-A)^2,N) A=12, K-A=13. Therefore m'=13*5, A(K-A)=13*12, (K-A)^2=13^2 and N=5*12. Thus m=56, N=60. .12.... 20..... ....... .1...25 .5..606 1.(Cb,Eb,b^2,CE) This is equivalent to C^2+E^2=b^2. C=20, b=10?, E=??. 400=E^2-b^2=200*2=100*4=50*8=40*10=20*20 Thus 400=101^2-99^2=52^2-48^2=29^2-21^2=25^2-15^2=20^2-0^2. b=101, E=99. .12.... 20...99 .1..... .1...25 .5..606 2.(Cd,A'd,d^2,j) This is equivalent to C^2+(A')^2=d^2 and C(A')=j. C=20, A'=21. Thus d=29, j=420. .12..2. 20...99 .1...4. .1...2. .5..606 3.(G,M(K+A)/(d-A),f,j) (?4?,5??(37/17),????,420) is of the form (xz,yz,z^2,xy). 420=xy, where x^2+y^2 is square and (x,y) are coprime. x^2+y^2>y^2. Therefore x^2+y^2>=(y+1)^2. Thus x^2>=(2*y+1), and similarly y^2>=(2*x+1). This shows that x>=12 and y>=12. Thus possibilities are {x,y}={12,35},{15,28},{20,21}. 15^2+28^2=1009 is not square. Thus <{x,y},z>=<{12,35},37>,<{20,21},29> f=z^2 is 4-digits. Thus f=1369, and the two parts are 12*37=444 and 35*37=1295. G is a 3-digit number with a central 4. Therefore G=444, and 37M/17=1295. Thus M=595. .12..2. 201..99 .13.444 .16..25 .595606 6.(M-j,N(A'-H),D(K-C),B'+M-J) M-j=595-420=175 175=xz where z^2-x^2 is square. z>x. Thus =<1,175> or <5,35> or <7,25> GCD(x,z)=1, getting rid of x=5. 175^2-1 is not square. Thus z=25, x=7, y=24. Thus 24*25=N(A'-H)=60*(21-H). Thus H=11. 25^2=D(K-C)=D(25-20). Thus D=125. 7*24=B'+M-J=B'+595-J. Thus J=427+B'. .12..2. 2012599 .13.444 116..25 .595606 5.((K+N)(m-C),c,Kg,H(J-G+A'-C)) (K+N)(m-C)=(25+60)(56-20)=85*36. 85*36=xz where z^2-x^2 is square and GCD(x,z)=1. If x<=18, then z>=170, and hence 2z-1>=339>324=x^2. Thus z^2-x^2>z^2-(2z-1)=(z-1)^2. Thus x>18. If x>=60, then z<=51. Hence x<60. Thus is <20,153>,<30,102>,<34,90>,<36,85>,<45,68>,<51,60>. GCD(x,z)=1. Hence is <20,153>,<36,85>,<45,68>. The only one with z^2-x^2 square is =<36,85>, with z^2-x^2=77^2. c=77*85=6545. Kg=85^2. K=25. Thus g=17^2=289. 36*77=H(J-G+A'-C)=11(695-444+21-20). From before B'=J-427=695-427=268. Thus B=862. .12862. 2012599 .138444 1169525 .595606 4.(J-a',(d-a')(h+k),N(F+e)/C+J-M,HN) a,F,h,e HN=11*60. 11*60=xy where x^2+y^2 is square and GCD(x,y)=1. {x,y} is {4,165},{12,55},{20,33},{44,15},{220,3},{132,5},{60,11},{660,1}. y^2>=2x+1 and x^2>=2y+1. Thus {x,y} is {12,55},{20,33},{44,15},{60,11}. The only one of these with x^2+y^2 is {x,y}={11,60} and z=61. J-a'=11*61 or 60*61. J=695. Therefore J-a'=11*61, and a'=24 60*61=(d-a')(h+k)=(29-24)(h+15). Thus h=717. 61*61=N(F+e)/C+J-M=60(713+e)/20+695-595. Thus e=494. 4128624 2012599 7138444 1169525 7595606 Luke Pebody