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 <x,z>=<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 <x,z> is <20,153>,<30,102>,<34,90>,<36,85>,<45,68>,<51,60>.
GCD(x,z)=1.
Hence <x,z> is <20,153>,<36,85>,<45,68>.
The only one with z^2-x^2 square is <x,z>=<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