Question 935274
Then {{{a/d=x}}} and {{{N/d=M}}} are integers.
{{{N/d=M}}}--->{{{N=dM}}}
{{{a/d=x}}}--->{{{a=dx}}}
{{{system(a=dx,N=abc+1)}}} ---> {{{N=dxbc+1}}}
{{{system(N=dM,N=dxbc+1)}}} ---> {{{dM=dxbc+1}}}--->{{{dM-dxbc=1}}}--->{{{d(M-xbc)=1}}}
Since {{{M}}} , {{{x}}}, {{{b}}}, and {{{c}}} are all integers, so is {{{M-xbc}}} ,
and since the product of integers {{{d}}} and {{{M-xbc}}} is {{{1}}} ,
they must both be {{{1}}} :
{{{M-xbc=1}}} and more importantly {{{d=1}}} .
 
The way that is proven for {{{a}}} ,
it can be proven for {{{b}}} and {{{c}}}
(but in a raesonable world it should not be required),
because they all play the same role with different names.