# Possible mathematical induction problem

• Dec 2nd 2013, 02:03 PM
Possible mathematical induction problem
Course: Foundations of Higher MAth

Prove that $24|(5^{2n} -1)$ for every positive integer n.

This is a question from my final exam today.

P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.

Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$

P(k+1): $24|(5^{2(k+1)}-1)$

$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$

Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?
• Dec 2nd 2013, 03:09 PM
Plato
Re: Possible mathematical induction problem
Quote:

Course: Foundations of Higher MAth
Prove that $24|(5^{2n} -1)$ for every positive integer n.
This is a question from my final exam today.
P(n): $24|(5^{2n}-1)$
P(1): $24|(5^2 -1)$ is a true statement.
Assume P(k) is true. Then $5^{2k}-1 = 24a$ for some integer a.
Then $5^{2k}= 24a + 1$
P(k+1): $24|(5^{2(k+1)}-1)$
$5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$
$=(24a + 1)*25 -1$
$=(24a)(25)+25 -1$
$=(24a)(25)+24$
$=24(25a+1)$
$=24b$
Therefore, $24|(5^{2(k+1)}-1)$. By PMI, P(n) is true for every positive integer n.
None of my friends used this method though. Is this a correct way to do it?

A strict grader may like to have seen more grouping symbols.
However, the argument is correct.