MHB Can $3^{2008}+4^{2009}$ Be Factored into Two Numbers Larger Than $2009^{182}$?

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The discussion centers around the expression $3^{2008}+4^{2009}$ and whether it can be factored into two positive integers, both exceeding $2009^{182}$. Participants express uncertainty about the solution but suggest that exploring properties of the numbers involved could lead to insights. Some propose starting points for analysis, emphasizing the need for a rigorous approach to prove the factorization. The conversation highlights the mathematical challenge and encourages collaborative problem-solving. Ultimately, the goal remains to determine the factorization of the expression in question.
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Show that $3^{2008}+4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$
 
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anemone said:
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
 
Thanks for the food for thought, kaliprasad!

Solution proposed by other:
We use the standard factorization:

$$x^4+4y^4=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$$

Observe that for any integers $x, y$,

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

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

We write

$$3^{2008}+4^{2009}=3^{2008}+4(4^{2008})$$

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

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

with both

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

and

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

And notice that

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

and hence we're done.
 
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