97yz-97z-97=19xyz equation A
here 97-z-1 <yz
or, yz-z-1 <yz
or, 97(yz-z-1)<97yz
or,19xyz <97yz
or, 19x <97
or, x<97/19
or, x=1,2,3,4,5 is equation B
gcd(19,97)=1
or,at least one of x,y,z is multiple of 97 not=x
let y=97m
or, 97(97mz-z-1)=19x97mz
or,(97mz-z-1)=19xmz
or,(97-19x)mz=z+1 is equation C
here (97-19x)>=2 since x=1,2,3,4,5
(97-19x)mz>=2mz>=2z>z+1
and we have a contradiction with C
or, y cannot be multiple of 97
then let z=97m then A
becomes 97(y97m-97m-1)=19xy97m
or,97ym-97m-1=19xym
or,(97y-97-19xy)m=1 is equation D
since these are positive integer quantities, D implies that both m and 97y-97-19xy=1, m=1 or, z=97
or, 97y-19-19xy=1
or,97y-19xy=98
or,(97-19x)y=98
or,y=98/(97-19x)
where from A we get 97-19x=78,59,0,21and 2 for x=0,1,2,3,4,5
since y is positive integer, it follows that the first four values not taken, x=5,97-19x=2,y=49,z=97.
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