Proof by Induction: For any integer $n > 0$, $3|(5^{2n} − 1)$.

Question

Is there any difference in proving "for any integer $n > 0$, $3|(5^{2n}-1)$" and proving "for any integer $n \ge 0$, $3|(5^{2n} -1)$"? Or are they the same?

Answer 1

The second requires that you prove it for $n=0$ while the first does not. If you want to prove the second, you could just say $5^{2\cdot 0}-1=0$, which is divisible by $3$ and then prove the first. You could also use this as your base case and do induction from there. If you only want to prove the first you could use $n=1$ as your base case and note that $5^{2\cdot 1}-1=24$, which is divisible by $3$.

Answer 2

Observe that when $n=0$, your statement holds true since 0 is divisible by 3. However, if you assume $n>0$, then you have to show that when $n=1$, the statement also holds true which in this case it is true since 24 is also divisible by 3. Either way the end result is the same.

Next, we assume the statement holds true for $n=k$, that is $5^{2k}-1$ is divisible by $3$. We will show that the statement also holds true for $n=k+1$. Now, observe that $$5^{2(k+1)}-1 = 25\, .\,5^{2k}-1 = 25\, .\,5^{2k}-1 -25+25= 25(5^{2k}-1)+24 $$ By our hypothesis of $n=k$, $5^{2k}-1$ is divisible by $3$. Moreover, $24$ is also divisible by $3$. Therefore $25(5^{2k}-1)+24$ is also divisible by $3$.

Thus, by assuming $n=k$ is true, we obtain $n=k+1$ is also true. This completes the proof of induction for this problem.

Answer 3

Since the logical niceties of the distinction 'twixt the $n \ge 0$ and the $n > 0$ cases have been more than adequately analyzed by our colleagues Ross Milliken and Evan William Chandra, I'll cut to the chase and give a short inductive proof that $3 \mid (5^{2n} - 1)$ when $n \ge 1$.

Factoid: If $n \ge 1$ then $3$ divides exactly one of $n$, $n + 1$, $n + 2$.

Proof of Factoid: For $n = 1$, $n, n + 1, n + 2 = 1, 2, 3$, so the Factoid binds and we may take this as the base case. Now suppose $3$ divides exactly one of $k, k + 1, k + 2$; considering $k + 1, k + 2, k + 3$, if $3 \not \mid k + 1$, $3 \not \mid k + 2$, then by assumption $3 \mid k$, so $3 \mid k + 3$, and thus $3$ divides exactly one of $k + 1$, $k + 2$, $k + 3$. End: Proof of Factoid.

We apply this to the question of whether $3 \mid 5^{2n} - 1$; observe that $5^{2n} - 1 = (5^n - 1)(5^n + 1)$; by our Factoid, $3$ divides exactly one of $5^n - 1$, $5^n$, $5^n + 1$; if $3 \mid 5^n$, then $3 \mid 5$ since $3$ is prime; but this is clearly false, so $3 \mid 5^n - 1$ or $3 \mid 5^n + 1$; then $3 \mid (5^n - 1)(5^n + 1) = 5^{2n} - 1$.

And that's that!