- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,909

- Jan 25, 2013

- 1,225

$\large 3^{4^5}=3^{1024}$

$\large 4^{5^6}=4^{15625}=(3+1)^{1024}\times 4^{14601}$

using binomial expansion the first part is done

Last edited:

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

. . [tex]3^{4^5} \;=\;3^{1024}[/tex]

. . [tex]4^{5^6} \;=\;4^{15,625}[/tex]

However: .[tex](3^4)^5 \;=\;8^4\:\text{ and }\: (4^5)^6 \:=\:1024^6[/tex]

- Jan 25, 2013

- 1,225

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

now the solution is as follows :

$(3^{512})^2+(2^{15625})^2---(1)$

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

now the solution is as follows :

$(3^{512})^2+(2^{15625})^2---(1)$

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

Last edited:

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,909

HiHello, albert!

You are misreading the exponents.

In an exponential "stack",

. . we read.from the top down

. . [tex]3^{4^5} \;=\;3^{1024}[/tex]

. . [tex]4^{5^6} \;=\;4^{15,625}[/tex]

However: .[tex](3^4)^5 \;=\;8^4\:\text{ and }\: (4^5)^6 \:=\:1024^6[/tex]

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

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

now the solution is as follows :

$(3^{512})^2+(2^{15625})^2---(1)$

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

Suggested solution by Pedro and Alex:

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

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.