Number theory- mathematical induction?

Question

use mathematical induction to prove that 21 divides 4^n+1 + 5^2n-1 for n=1,2,3......
Use headings base case and induction?

So letting p(n) be the proposition that 21 divides 4^n+1 + 5^2n-1
I can prove the base case is true, that is p(1), where n=1 is true since 4^2 + 5^1= 21 and clearly 21 divides 21
I am struggling with induction, that is, assuming p(k) is true, where p(k)=4^k+1 + 5^2k-1, and proving from that p(k+1)???

Answer 1

Actually, p(k) is the STATEMENT that 21 divides 4^(k+1)+5^(2k-1). So we assume this part. Now we're off to show that 21 divides 4^(k+2)+5^(2k+1), which is p(k+1). Well,
4^(k+2) + 5^(2k+1)
= 4*4^(k+1) + 25*5^(2k-1)
= 4*4^(k+1) + 4*5^(2k-1) + 21*5^(2k-1)
= 4*(4^(k+1) + 5^(2k-1)) + 21*5^(2k-1)

Now from our induction hypothesis (i.e., our assumption that p(k) is true), the first term (in parentheses) is divisible by 21. But the third term is clearly also divisible by 21, therefore the whole sum is, and we have proved p(k+1).

So, by induction, the statement holds for all n=1,2,3,...

Answer 2

Let n = 1, 4^2 + 5^1 = 21 is divisible by 21
Let us assume that 4^(n+1) + 5^(2n-1) is divisible by 21. It is sufficient to demonstrate that 4^(n+2) + 5^(2n +1) is also divisible by 21.
Let us consider the divisibility of 4^(n+2) + 5^(2n +1) by both 3 and 7 when 4^(n+1) + 5^(2n-1) is divisible by both 3 and 7.
Since 4^(n+1) ==1 mod 3, we must have 5^(2n-1) ==2 mod 3. Hence 5^(2n +1) mod 3 = 5^(2n-1) *5^2 mod 3 == 2*2^2 mod 3 == 2 mod 3. Also 4^(n+2) ==1 mod 3. Therefore, 4^(n+2) + 5^(2n +1) ==0 mod 3.
Since 4^(n+1) ==4, 2, or 1 mod 7, we must have 5^(2n-1) ==3, 5, or 6 mod 7, correspondingly. We then discuss these 3 cases, and it is easy to demonstrate that 4^(n+2) + 5^(2n +1) mod 7 == 4^(n+1)*4 + 5^(2n-1) *5^2 mod 7 == 0 mod 7.
Therefore, we completed the induction.