SOLUTION: Prove that 5^{3^n} + 1 is divisible by 3^{n + 1} for all nonnegative integers n. Prove this using induction. Can you please explain in detail? Thank you so much!

Question 1177009: Prove that 5^{3^n} + 1 is divisible by 3^{n + 1} for all nonnegative integers n. Prove this using induction. Can you please explain in detail? Thank you so much!
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Absolutely, let's prove that $5^{3^n} + 1$ is divisible by $3^{n+1}$ for all nonnegative integers $n$ using the Lifting The Exponent Lemma (LTE).
**Understanding the Problem**
We want to show that $3^{n+1} \mid (5^{3^n} + 1)$ for all $n \geq 0$.
**Lifting The Exponent Lemma (LTE)**
The LTE Lemma states that if $x$ and $y$ are integers, $n$ is a positive integer, and $p$ is an odd prime such that $p \mid (x+y)$ but $p \nmid x$ and $p \nmid y$, then:
$v_p(x^n + y^n) = v_p(x+y) + v_p(n)$ if $n$ is odd.
Here, $v_p(m)$ denotes the highest power of $p$ that divides $m$.
**Applying LTE**
1. **Base Case (n = 0):**
When $n=0$, we have $5^{3^0} + 1 = 5^1 + 1 = 6$. We need to show that $3^{0+1} = 3^1 = 3$ divides $6$. Clearly, $3 \mid 6$, so the base case holds.
2. **Inductive Step:**
Let's assume that $3^{n+1} \mid (5^{3^n} + 1)$ for some nonnegative integer $n$. We want to show that $3^{n+2} \mid (5^{3^{n+1}} + 1)$.
We have $5^{3^{n+1}} + 1 = (5^{3^n})^3 + 1^3$.
Let $x = 5^{3^n}$ and $y = 1$. Then we have $x^3 + y^3$.
We know that $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.
So, $5^{3^{n+1}} + 1 = (5^{3^n} + 1)((5^{3^n})^2 - 5^{3^n} + 1)$.
We want to find $v_3(5^{3^{n+1}} + 1)$.
We can apply LTE to $x^3 + y^3$:
$v_3(5^{3^{n+1}} + 1) = v_3((5^{3^n})^3 + 1^3) = v_3(5^{3^n} + 1) + v_3(3)$.
By the inductive hypothesis, $3^{n+1} \mid (5^{3^n} + 1)$. Therefore, $v_3(5^{3^n} + 1) = n+1$.
Also, $v_3(3) = 1$.
Thus, $v_3(5^{3^{n+1}} + 1) = (n+1) + 1 = n+2$.
This means that $3^{n+2} \mid (5^{3^{n+1}} + 1)$.
**Conclusion**
By the principle of mathematical induction, $5^{3^n} + 1$ is divisible by $3^{n+1}$ for all nonnegative integers $n$.
**Explanation of LTE Application**
* We identified $x = 5^{3^n}$ and $y = 1$.
* We observed that $3 \mid (5^{3^n} + 1)$ by the inductive hypothesis.
* We noted that $3 \nmid 5^{3^n}$ and $3 \nmid 1$.
* We applied the LTE formula for odd $n=3$, which gave us $v_3(x^3 + y^3) = v_3(x+y) + v_3(3)$.
* We used the inductive assumption to replace $v_3(5^{3^n} + 1)$ with $n+1$.
* We used the fact that $v_3(3) = 1$.
This allowed us to conclude that $v_3(5^{3^{n+1}} + 1) = n+2$, which implies $3^{n+2} \mid (5^{3^{n+1}} + 1)$.

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