Using algebraic identities to prove a number is not prime

Question

$$3^{3^n}(3^{3^n}+1)+3^{3^n +1}-1$$ I want the prove that the number is not prime. I used the identity $a^3+b^3+c^3-3abc $. I couldn't simplify to the state where the factors could be observed.

Answer 1

The solution suggested that $$a=3^{3^{n-1}} , b=9^{3^{n-1}}, c=-1$$ I can't get it. How could I derive those terms from the equation. Then, followed by the identity $$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

Answer 2

define $$ w = 3^{3^n} $$

Your number is then $$ w(w+1) + 3 w - 1 = w^2 +w + 3w -1 = w^2 + 4 w -1 $$

Now, $w$ is odd, so $w^2$ is odd, so $w^2 + 4 w -1$ is even. Also, $w \geq 3,$ so $w^2 + 4 w -1$ is bigger than 4 and even, therefore not prime.

Answer 3

The other approach does show a quite large factor of the number; good to know how to approach this.

The expression $$ 3^{3^n} 3^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$ first becomes $$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$

The first two terms really are cubes, since the exponents are divisible by 3. If we look for $a^3 = 3^{3^n},$ it is reasonable to try $$ a = 3^x $$ because we want a power of three. Then $$ a^3 = \left(3^x \right)^3 = 3^{3x}. $$ Then $a^3 = 3^{3^n}$ tells us $$ 3^{3x} = 3^{3^n} $$ so $$ 3x = 3^n $$ and $$ x = 3^n 3^{-1} = 3^{n-1}, $$ $$ a = 3^x = 3^{3^{n-1}} $$ $$ $$ If we look for $b^3 = 9^{3^n},$ it is reasonable to try $$ b = 9^y $$ because we want a power of nine. Then $$ b^3 = \left(9^y \right)^3 = 9^{3y}. $$ Then $b^3 = 9^{3^n}$ tells us $$ 9^{3y} = 9^{3^n} $$ so $$ 3y = 3^n $$ and $$ y = 3^n 3^{-1} = 3^{n-1}, $$ $$ b = 9^x = 9^{3^{n-1}} $$ $$ $$ We are not yet sure what we want to use for $c.$ Let us examine what we already have for $3ab,$ anticipating the eventual $-3abc.$

$$ a b = 3^{3^{n-1}} 9^{3^{n-1}} $$

MORE TO COME 11:08 am

We do need to understand $b = a^2.$ Well, $$ a^2 = \left( 3^{3^{n-1}}\right)^2 = 3^{ 2 \cdot 3^{n-1}} $$ while $$ b = \left( 3^2 \right)^{3^{n-1}} = 3^{ 2 \cdot 3^{n-1}} \; . $$ So $b = a^2,$ also $ab = a a^2 = a^3$ so $$ ab = 3^{3^n} \; . $$ Next $$ 3ab = 3 \cdot 3^{3^n} = 3^1 \cdot 3^{3^n} = \cdot 3^{3^n + 1} \; .$$ This is progress. We now have the overall expression $$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 + 3ab - 1. $$ We can introduce a value of $c$ by demanding $-3abc = 3ab.$ The requires $c=-1.$ We have $$ b^3 + a^3 - 3abc - 1. $$ Finally $c^3 = (-1)^3 = -1,$ and we get $$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 = b^3 + a^3 - 3abc - c^3 = a^3 + b^3 + c^3 - 3abc. $$

Perhaps I should say why the new factors are both large. We have $$a+b+c > b = 9^{3^{n-1}}$$ It is also worth remembering that $$ a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right) $$ which is bigger than $ (b+1)^2 / 2 \; > \; b^2 / 2 \; . $