# Proofs

• Mar 24th 2009, 12:00 PM
dlee426
Proofs
I need to show that 2^n + 3^n is a multiple of 5 where n is an odd number using mathematical induction.

I am pretty far along the problem, having done the induction, calling m=2k+1, then plugging in m+2=2k+3, but I cannot manipulate the problem to show that it is a multiple of 5.

• Mar 24th 2009, 12:16 PM
Moo
Hello,
Quote:

Originally Posted by dlee426
I need to show that 2^n + 3^n is a multiple of 5 where n is an odd number using mathematical induction.

I am pretty far along the problem, having done the induction, calling m=2k+1, then plugging in m+2=2k+3, but I cannot manipulate the problem to show that it is a multiple of 5.

So you want to prove that $\displaystyle 2^{2k+3}+3^{2k+3}$ is a multiple of 5, knowing that $\displaystyle 2^{2k+1}+3^{2k+1}=5n$, for some integer n (inductive hypothesis)
Note that $\displaystyle 3^{2k+1}=5n-2^{2k+1}$

Hence :
\displaystyle \begin{aligned} 2^{2k+3}+3^{2k+3} &=4 \cdot 2^{2k+1}+9 \cdot 3^{2k+1} \\ &=4 \cdot 2^{2k+1}+9 \cdot (5n-2^{2k+1}) \\ &=4 \cdot 2^{2k+1}+5 \cdot (9n)-9 \cdot 2^{2k+1} \\ &=2^{2k+1} (4-9)+5 \cdot (9n) \end{aligned}

... :)
• Mar 24th 2009, 01:09 PM
gosualite
Without induction, consider that $\displaystyle 3 \equiv -2 \mod 5$, so $\displaystyle 2^n+3^n \equiv 2^n+(-2)^n \mod 5$. For odd values of n, you can factor out the negative, giving what you want.