Possible mathematical induction problem

Course: Foundations of Higher MAth

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

This is a question from my final exam today.

P(n): $\displaystyle 24|(5^{2n}-1)$

P(1): $\displaystyle 24|(5^2 -1)$ is a true statement.

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

Then $\displaystyle 5^{2k}= 24a + 1$

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

$\displaystyle 5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$

$\displaystyle =(24a + 1)*25 -1$

$\displaystyle =(24a)(25)+25 -1$

$\displaystyle =(24a)(25)+24$

$\displaystyle =24(25a+1)$

$\displaystyle =24b$

Therefore, $\displaystyle 24|(5^{2(k+1)}-1)$. 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?

Re: Possible mathematical induction problem

Quote:

Originally Posted by

**MadSoulz** Course: Foundations of Higher MAth

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

This is a question from my final exam today.

P(n): $\displaystyle 24|(5^{2n}-1)$

P(1): $\displaystyle 24|(5^2 -1)$ is a true statement.

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

Then $\displaystyle 5^{2k}= 24a + 1$

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

$\displaystyle 5^{2(k+1)}-1=5^{2k+2}-1=5^{2k}*5^2-1$

$\displaystyle =(24a + 1)*25 -1$

$\displaystyle =(24a)(25)+25 -1$

$\displaystyle =(24a)(25)+24$

$\displaystyle =24(25a+1)$

$\displaystyle =24b$

Therefore, $\displaystyle 24|(5^{2(k+1)}-1)$. 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.