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Sum to Product Challenge

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anemone

MHB POTW Director
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Feb 14, 2012
3,683
Show that $3^{2008}+4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$
 

kaliprasad

Well-known member
Mar 31, 2013
1,309
Show that $3^{2008}+4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$
I do not know the solution but below could be a starting point

(x^4+ 4y^2) = (x^2 + 2y^2 – 2xy)(x^2 + 2y^2 + 2xy)

SO 3^2008+ 4^ 2009 = (3^502)^4 + 4 * (4^502)^ 4
= (3^1004 + 2 *4^1004 + 2 * 12^502) (3^1004 + 2*4^1004 - 2 * 12^502)

Now if we show that (3^1004 + 2*4^1004 - 2 * 12^502) > 2009^182 we are through
 
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anemone

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Feb 14, 2012
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Thanks for the food for thought, kaliprasad!

Solution proposed by other:

We use the standard factorization:

\(\displaystyle x^4+4y^4=(x^2+2xy+2y^2)(x^2-2xy+2y^2)\)

Observe that for any integers $x, y$,

\(\displaystyle x^2+2xy+2y^2=(x+y)^2+y^2 \ge y^2\) and

\(\displaystyle x^2-2xy+2y^2=(x-y)^2+y^2 \ge y^2\)

We write

\(\displaystyle 3^{2008}+4^{2009}=3^{2008}+4(4^{2008})\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(3^{502})^4+4(4^{502})^4\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=((3^{502})^2)^2+2((3^{502})^2(4^{502})^2+2((4^{502})^2)((3^{502})^2)^2-2((3^{502})^2(4^{502})^2+2((4^{502})^2)\)

with both

\(\displaystyle ((3^{502})^2)^2+2((3^{502})^2(4^{502})^2+2((4^{502})^2) \ge (4^{502})^2\)

and

\(\displaystyle ((3^{502})^2)^2-2((3^{502})^2(4^{502})^2+2((4^{502})^2) \ge (4^{502})^2\)

And notice that

\(\displaystyle (4^{502})^2=2^{2008}>2^{2002}=(2^{11})^{182}=2048^{182}>2009^{182}\)

and hence we're done.