Final Solution 13 Maria Acevedo(Recitation 2/21)

13. Statement: 52n+1+22n+1 is divisible by 7 for all n Є $\mathbb{N}$.
Proof: We can begin by proving that 52n+1+22n+1 being divisible by 7 is true for the base case, n = 0.
52(0)+1+22(0)+1 =
51+21=
5+2=7
7/7 =1
Based on this idea, assume the argument holds true for n=k.
52k+1+22k+1=7J where JЄ$\mathbb{N}$
22k+1=7J-52k+1
Next, we must show that the argument holds for n=k+1.
52(k+1)+1+22(k+1)+1=52k+3+22k+3
=52k+1(52)+22k+1(22)
=25(52k+1)+4(22k+1)
=25(52k+1)+4(7J-52k+1)
=25(52k+1)+28J-4(52k+1)
=21(52k+1)+28J
=7[3(52k+1)+4J]
7[3(52k+1)+4J] will always be divisible by 7 regardless of the value of k as long as k Є $\mathbb{N}$. Hence, 52n+1+22n+1 is divisible by 7 for all n Є N. ∎

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