Is the Product of Two Integers Greater Than 10^2009? Let's Prove It!

[anemone]

Prove that $\large 3^{4^5}+4^{5^6}$ is the product of two integers, each at least $\large 10^{2009}$.

 

[Albert1]

Prove that $\large 3^{4^5}+4^{5^6}$ is the product of two integers, each at least $\large 10^{2009}$.

Spoiler

$\large 3^{4^5}=3^{1024}$
$\large 4^{5^6}=4^{15625}=(3+1)^{1024}\times 4^{14601}$
using binomial expansion the first part is done

 

[soroban]

Hello, Albert!

Prove that $3^{4^5}+4^{5^6}$ is the product of two integers,
each at least $\large 10^{2009}$.

You are misreading the exponents.

In an exponential "stack",
. . we read from the top down.

. . $$3^{4^5} \;=\;3^{1024}$$

. . $$4^{5^6} \;=\;4^{15,625}$$However: .$$(3^4)^5 \;=\;8^4\:\text{ and }\: (4^5)^6 \:=\:1024^6$$

 

[Albert1]

thanks soroban , in a haste I made a mistake in misreading the exponent

now the solution is as follows :

Spoiler

$(3^{512})^2+(2^{15625})^2---(1)$
let $a=3^{512}, b=2^{15625}$
(1) becomes $(a+b)^2 -2ab=(a+b)^2 -[(3^{256}\times 2^{7813})]^2=(x+y)(x-y)$
$here \,\, x=a+b, y=(3^{256}\times 2^{7813}) $
the rest is easy:
we only have to compare x-y and $10^{2009}$(compare digit numbers of both values)
I only count them roughly
$15625\times log 2>15625\times 0.3>4687>2009$
$256\times log 3+7813\times log 2>102+2343$
4687-102-2343=2242>2009
$\therefore x-y >10^{2009}$
and the proof is finished

 

[anemone]

Hello, Albert!


You are misreading the exponents.

In an exponential "stack",
. . we read from the top down.

. . $$3^{4^5} \;=\;3^{1024}$$

. . $$4^{5^6} \;=\;4^{15,625}$$However: .$$(3^4)^5 \;=\;8^4\:\text{ and }\: (4^5)^6 \:=\:1024^6$$

Hi soroban,

Thanks for helping me to let Albert know that a misreading has occurred and that he has the chance to fix things right.:)

thanks soroban , in a haste I made a mistake in misreading the exponent

now the solution is as follows :

Spoiler

$(3^{512})^2+(2^{15625})^2---(1)$
let $a=3^{512}, b=2^{15625}$
(1) becomes $(a+b)^2 -2ab=(a+b)^2 -[(3^{256}\times 2^{7813})]^2=(x+y)(x-y)$
$here \,\, x=a+b, y=(3^{256}\times 2^{7813}) $
the rest is easy:
we only have to compare x-y and $10^{2009}$(compare digit numbers of both values)
I only count them roughly
$15625\times log 2>15625\times 0.3>4687>2009$
$256\times log 3+7813\times log 2>102+2343$
4687-102-2343=2242>2009
$\therefore x-y >10^{2009}$
and the proof is finished

Thanks Albert for participating and your solution as well!

Suggested solution by Pedro and Alex:

Spoiler

Let $m=3^{256}$ and $k=4^{3906}$. Then

$\begin{align*}\large 3^{4^5}+4^{5^6}&=m^4+4k^4\\&=(m^4+4m^2k^2+4k^4)-4m^2k^2\\&=(m^2+2mk+2k^2)(m^2-2mk+2k^2)\end{align*}$

Notice that $m^2+2mk+2k^2>m^2-2mk+2k^2>2k^2-2mk=2k(k-m)>k>2^{7800}>(10^3)^{780}>10^{2009}$.

The result is then follows.