Can $3^{2008}+4^{2009}$ be written as a product of two positive integers?

In summary, "writing a sum to a product" is the process of simplifying a mathematical equation by expressing the sum of two or more terms as a product. This can be done using the distributive property of multiplication. It is important because it helps to organize and simplify equations, making them easier to solve and manipulate. An example of writing a sum to a product is distributing a number outside of parentheses to each term inside, resulting in a product. As a scientist, this technique can be useful in simplifying complex equations and making them more manageable.
  • #1
anemone
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Show that $3^{2008}+4^{2009}$ can be written as a product of two positive integers each of which is larger than $2009^{182}$.
 
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  • #2
we have $3^{2008} + 4^{2009} = (3^{1004})^2 + (2^{2009})^2$
$= (3^{1004})^2 + (2^{2009})^2 + 2 * 3^{1004} * 2^{2009} - 2 * 3^{1004} * 2^{2009}$
$= ((3^{1004}) + (2^{2009}))^2 - 3^{1004} * 2^{2010}$
$= (3^{1004} + 2^{2009})^2 - (3^{502} * 2^{1005})^2$
$= (3^{1004} + 2^{2009} +3^{502} * 2^{1005}) (3^{1004} + 2^{2009} -3^{502} * 2^{1005})$

Let us consider the lower value
$(3^{1004} + 2^{2009} -3^{502} * 2^{1005})$
$=2^{2009} + 3^{502}(3^{502} - 2^{503})$

as $3^{502} = 9^{\frac{502}{2}} = 9^{251} > 8^{251} = 2^{251 * 3} = 2^{753} > 2^{502}$

so
$(3^{1004} + 2^{2009} -3^{502} * 2^{1005}) > 2^{[2009} = 2^{ 11 * 182+7} > (2^{11})^{182} = 2048 ^{182} > 2009 ^ {182}$

as lower term is $>2009^{182}$ so we are done
 
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1. Can $3^{2008}+4^{2009}$ be written as a product of two positive integers?

Yes, it can be written as a product of two positive integers.

2. What are the factors of $3^{2008}+4^{2009}$?

The factors of $3^{2008}+4^{2009}$ are $3^{2008}$ and $4^{2009}$.

3. Is $3^{2008}+4^{2009}$ a prime number?

No, $3^{2008}+4^{2009}$ is not a prime number as it can be factored into two positive integers.

4. Can $3^{2008}+4^{2009}$ be written as a power of a single number?

No, $3^{2008}+4^{2009}$ cannot be written as a power of a single number as it is the sum of two distinct powers.

5. Is there a general formula for finding the factors of $3^{n}+4^{n}$?

Yes, the factors of $3^{n}+4^{n}$ can be expressed as $(3^{k}+4^{k})(3^{n-k}-4^{n-k})$ where $k$ is any integer from $0$ to $n$.

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