Question 252126
<font size = 8 color = "red"><b>Note: The tutor "drk" did not show that
3^2008[1 +3*(4/3)^2009] is the product of two integers.  In fact
[1 +3*(4/3)^2009] is not an integer!!!</font></b>


show that 3^2008+4^2009 can be written as product of two positive integers each of which is larger than 2009^182.
<pre><font size = 4 color = "indigo"><b>
We use the fact that

{{{3^2008+4^2009=3^2008+4*4*2008=(3^502)^4+4(4^502)}}}

We let {{{x=3^502}}} and {{{y =4^502}}}

Then

{{{(3^502)^4+4(4^502)^4=x^4+4y^4=x^4+red(4x^2y^2)+4y^4-red(4x^2y^2)=(x^4+4x^2y^2+y^4)-4x^2y^2}}}
{{{""=(x^2+2y^2)^2-(2xy)^2=((x^2+2y^2)-2xy)((x^2+2y^2)-2xy)=(x^2-2xy+2y^2)(x^2+2xy+2y^2)}}}  

Substituting back for {{{x}}} and {{{y}}}

{{{(x^2-2xy+2y^2)(x^2+2xy+2y^2)}}}

{{{((3^502)^2-2(3^502)(4^502)+2(4^502)^2)((3^502)^2+2(3^502)(4^502)+2(4^502)^2)}}}
  
{{{""=(3^1004-2*3^502*4^502+2*4^1004)(3^1004+2*3^502*4^502+2*4^1004)}}}

So this is the product of two very large integers. So we have

shown the first part.

Now we only need to show that the smaller of these, which is

{{{3^1004-2*3^502*4^502+2*4^1004}}}

is greater than {{{2009^182}}}.

I have figured out how to do this yet, but I'm working on it.

If and when I get it then I'll post it here.

Edwin</pre>