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RE: Challenge problem


In[11]:=
Together[ 3q^3 + 10q^2 + 3q  /. q -> a/b ]

 <<Picture (Metafile)>>

In[13]:=
Simplify[ % /. a -> j b ]

Demonstrates that is a is an integer times b, then each of the terms and
therefore the sum are integers. It doesn't prove that there are not
other solutions - not sure if there are.

-----Original Message-----
From: Coleman, Mark [mailto:Mark.Coleman at LibertyMutual.com]
To: mathgroup at smc.vnet.net
Subject: [mg71213] [mg71123] Challenge problem

I recently received the annual newsletter from the Math-Stats department
of my undergraduate alma mater. In part of the newsletter they posed the
following challenge problem:

"For which rational numbers q is 3q^3 + 10q^2 + 3q an integer?"

The problem comes from an annual mathematics competition the school
sponsors.  I fumbled around a bit, using Mathematica v5.2 to attempt an
answer, but without much luck. Of course I'm a statistician, not an
algebraist :-). But my curiosity is now piqued and I was wondering if
someone might outline an elegant answer using Mathematica.

Thanks,

-Mark


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