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

In summary, the Sum to Product Challenge is a mathematical problem where two numbers must be found that add up to a given sum and multiply to a given product. Its purpose is to exercise mathematical thinking and problem-solving skills. Some strategies for solving the challenge include listing factors and using algebraic equations. There can be multiple solutions to the challenge, and it has real-life applications in areas such as cryptography and engineering.
  • #1
anemone
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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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  • #2
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
 
  • #3
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.
 

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

What is the Sum to Product Challenge?

The Sum to Product Challenge is a mathematical problem where the goal is to find two numbers that when added together, will equal a given sum, and when multiplied together, will equal a given product.

What is the purpose of the Sum to Product Challenge?

The purpose of the Sum to Product Challenge is to exercise mathematical thinking and problem-solving skills. It is often used as a brain teaser or puzzle to challenge individuals' understanding of numbers and their relationships.

What are some strategies for solving the Sum to Product Challenge?

One strategy for solving the Sum to Product Challenge is to start by listing all possible factors of the given product and then pairing them up to see if any of the pairs add up to the given sum. Another strategy is to use algebraic equations to represent the given sum and product and solve for the unknown numbers.

Is there only one solution to the Sum to Product Challenge?

No, there can be multiple solutions to the Sum to Product Challenge. For example, if the given sum is 10 and the given product is 24, the solutions could be 6 and 4 or 8 and 3. It is important to check all possible combinations to ensure all solutions are found.

What are some real-life applications of the Sum to Product Challenge?

The Sum to Product Challenge has many real-life applications, including in cryptography where it is used to create secure codes and in engineering where it is used to calculate the resistance of circuits. It can also be used in everyday situations, such as finding the dimensions of a rectangle with a given perimeter and area.

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