Originally Posted by

**dwsmith** You are correct about saying d is 1 and p but I beg to differ on the rest.

Doesn't follow from above?

$\displaystyle d(m+r)=dq$ where q=m+r. Now, $\displaystyle dq=[n+(n+p)]\Rightarrow d|[n+(n+p)]\Rightarrow d|n \ \mbox{and} \ d|(n+p)$

No, by all means no!! It isn't true at all that $\displaystyle d\mid [n+(n+p)]\Longrightarrow d\mid n\,\,and\,\,d\mid (n+p)$ .

For example, take $\displaystyle d=3\,,\,n=1\,,\,p=19.\,\,Then\,\,3\mid 1+(1+19)\,,\,\,but\,\,3\nmid 1\,\,and\,\,3\nmid (1+19)$ !

Clearly, if d is 1, it is true sinc $\displaystyle 1|n\Rightarrow 1*k=n, \ k=n \ \mbox{and} \ 1*t=n+p, \ t=n+p$

Thus, 1 is a trivial case since it always works.

Now, examining $\displaystyle d|(n+p)\Rightarrow d|n \ \mbox{and} \ d|p$

Again, this doesn't follow!

I think you must check your logical deductions in this matter very carefully.

Tonio

which leads d=1 or p since 1 or the prime number divides p.

Also, it does add value. Not everyone thinks the same way and I don't prefer Melese proof. That doesn't mean I discredit his work. I appreciate it his work since it has benefited me in the past.