# Possible mathematical induction problem

• Dec 2nd 2013, 02:03 PM
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?
• Dec 2nd 2013, 03:09 PM
Plato
Re: Possible mathematical induction problem
Quote:

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.