Show that α=³√(7+5√2)+³√(7-5√2) is an integer. No calculators or computers allowed!

Wow, I didn't think I could do it, but I did. I just hoped that I was going down the right path and that things would become evident. Ok, here goes:

I will call A = 7, and B = 5√2 for simplification until the end.

a = ³√{A+B} + ³√{A-B}

a^3 = (³√{A+B} + ³√{A-B})^3
= (³√{A+B} + ³√{A-B}) * (³√{A+B} + ³√{A-B}) * (³√{A+B} + ³√{A-B})
=[ ³√{A+B}^2 + 2³√{A+B}³√{A-B} + ³√{A-B}^2 ] * (³√{A+B} + ³√{A-B})
= ³√{A+B}^3 + ³√{A+B}^2*³√{A-B} + 2³√{A+B}^2*³√{A-B} + 2³√{A+B}³√{A-B}^2 + ³√{A+B}³√{A-B}^2 + ³√{A-B}^3
= ³√{A+B}^3 + 3³√{A+B}^2*³√{A-B} + 3³√{A+B}³√{A-B}^2 + ³√{A-B}^3
= (A+B) + 3³√{A+B}^2*³√{A-B} + 3³√{A+B}³√{A-B}^2 + (A-B)
= 2A + 3³√{A+B}^2*³√{A-B} + 3³√{A+B}³√{A-B}^2
= 2A + 3³√{(A+B)^2*(A-B)} + 3³√{(A+B)(A-B)^2}

Ok, side note time: Notice that (A+B)^2 * (A-B) = (A+B)(A+B)(A-B) = (A+B)[A^2 – B^2]. This is just a simpler way of multiplying out those factors (less adding terms, less room for error). Similarly, (A+B) * (A-B)^2 = (A+B)(A-B)(A-B) = [A^2 – B^2](A-B). Ok, moving on

a^3 = 2A + 3³√{[A^2 – B^2](A+B)} + 3³√{[A^2 – B^2](A-B)}
= 2A + 3³√{A^3 + A^2*B – A*B^2 – B^3} + 3³√{A^3 – A^2*B – A*B^2 + B^3}

Ok, I think that's about as far as I can take it with the As and Bs. Now let's substitute in their values.

A^3 = 7^3 = 49*7 = 343 (done by hand!)
A^2*B = 7^2 * 5√2 = 49 * 5√2 = 245√2
A*B^2 = 7*(5√2)^2 = 7*(5^2 * √2^2) = 7 * 25 * 2 = 7*50 = 350
B^3 = (5√2)^3 = 5^3 * √2^3 = 125 * √2^2 * √2 = 125 * 2 * √2 = 250√2

Ok, so:

a^3 = 2A + 3³√{343 + 245√2– 350 – 250√2} + 3³√{343 – 245√2– 350 + 250√2}
= 2A + 3³√{ – 7 – 5√2 } + 3³√{– 7 + 5√2}
= 2A + 3³√{(-1)(7 + 5√2)} + 3³√{(-1)(7 - 5√2)}
= 2A + 3³√{(-1)}³√{7 + 5√2} + 3³√{(-1)}³√{7 - 5√2}

Well, the cube route of –1 is just –1. So we have

a^3 = 2A – 3³√{7 + 5√2} – 3³√{7 - 5√2}
= 2A – 3[³√{7 + 5√2} + ³√{7 - 5√2}]

But wait, the given is that a = ³√{7 + 5√2} + ³√{7 - 5√2}. So:

a^3 = 2A – 3a = 14 – 3a
a^3 + 3a = 14

This is nice and simple. Since we are dealing with small numbers I can probably just do the multiplications. I'll be able to see for what a value a^3 + 3a ends up being 14, using integers for a. Hopefully I won't find that for a certain number, a^3 + 3a < 14, and then when I increase a by one, get a^3 + 3a > 14. Since the problem asks "show that a is an integer" this means that a really should be an integer which means that I made a mistake somewhere. But if the problem had just asked "show if a is an integer or not" then that situation would have proven that there is not an integer. Enough rambling, on with the show:

I will ignore negative values for a. Why? Because if a<0 then a^3<0 and 3a <0, so the whole sum is <0. I am aiming for +14, so I'm not going to worry about negative numbers. So, starting at 1…

a=1: a^3 + 3a = 1^3 + 3(1) = 1+3 = 4
a=2: a^3 + 3a = 2^3 +3(2) = 8+6 = 14

Tada! a is indeed an integer. a=2. The end.

just cleaning up a bit

Edited on October 13, 2004, 1:43 pm

Posted by nikki on 2004-10-13 13:38:20