Linear, quadratic, module, parametric equations


Postby Guest » Mon Dec 09, 2019 2:41 pm

Solve in a set of nonnegative numbers

Re: x!+y!=z!

Postby HallsofIvy » Tue Dec 10, 2019 4:42 pm

I presume you mean "non-negative integers". It is easy to see that if x= y then x!+ y!= 2x!= z! has only x= y= 1, z= 2 as solution. If [tex]x\ne y[/tex] we can assume that x< y. In that case, y!= x!(x+1)(x+2)…(y-1)y= so that x!+ y!= x!(1+ (x+ 1)(x+ 2)…(y-1))= z! which has no solutions.

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