Monday, May 1, 2017

discrete mathematics - Prove using mathematical induction: for $n ge 1, 5^{2n} - 4^{2n}$ is divisible by $9$


I have to prove the following statement using mathematical induction.



For all integers, $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by 9.


I got the base case which is if $n = 1$ and when you plug it in to the equation above you get 9 and 9 is divisible by 9.


Now the inductive step is where I'm stuck.


I got the inductive hypothesis which is $ 5^{2k} - 4^{2k}$


Now if P(k) is true than P(k+1) must be true. $ 5^{2(k+1)} - 4^{2(k+1)}$


These are the step I gotten so far until I get stuck:


$$ 5^{2k+2} - 4^{2k+2} $$ $$ = 5^{2k}\cdot 5^{2} - 4^{2k} \cdot 4{^2} $$ $$ = 5^{2k}\cdot 25 - 4^{2k} \cdot 16 $$


Now after this I have no idea what to do. Any help is appreciated.


Answer



You're very close. Now add and subtract $4^{2k}$ in the first term to obtain $$ 5^{2k}\cdot 25-4^{2k}\cdot 16=25\cdot (5^{2k}-4^{2k})+(25-16)\cdot 4^{2k}=25\cdot (5^{2k}-4^{2k})+9\cdot 4^{2k} $$



The first term is divisible by $9$ by the induction hypothesis, hence the whole expression is divisible by $9$.


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