It took me several weeks to come up with this, hehe
Proof: Show that there is an integer x, such that nd=bcx.
By hypothesis,
we have an integer i, such that n=bi, and
we have an integer j, such that n=cj.
Since (b,c)=d, we have d|b and d|c by the definition of greatest common divisor.
Since d|b, we have k, such that b=dk.
Since d|c, we have h, such that c=dh
Note, that (h,k)=(b/d,c/d)=1, that is, h and k are relatively prime. [this step by a theorem about gcd()]
So, we have (dh)j=cj=n=bi=(dk)i by above hypothesis.
Hence, dhj=dki, and d!=0, we have hj=ki, hence h|ki
Since (h,k)=1, and h|ki, we have h|i [by the fundamental theorem of arithmetic].
Ok, since h|i , we can multiply both sides by an integer (dkd),
so dkdh|dkdi [by properties of divides | ]
(dk)(dh)|(dki)d, and since dk=b, dh=c, dki=n (see above), we have
bc|nd, as desired
I'll prove the comments in [ ... ] tomorrow. Or I can leave them as an exercise for the reader [EG] :)
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