Answer: We can prove that 5^(2n+1) + 2^(2n+1) is divisible by 7 for all n ∈ N (i.e., for all positive integers n) using mathematical induction.
Base case: When n = 1, we have:
5^(2n+1) + 2^(2n+1) = 5^(2(1)+1) + 2^(2(1)+1) = 5^3 + 2^3 = 125 + 8 = 133
133 is clearly divisible by 7, so the statement is true for n = 1.
Inductive step: Assume that the statement is true for some arbitrary positive integer k, i.e., assume that 5^(2k+1) + 2^(2k+1) is divisible by 7. We want to show that the statement is also true for k+1, i.e., that 5^(2(k+1)+1) + 2^(2(k+1)+1) is divisible by 7.
Using the laws of exponents, we can simplify 5^(2(k+1)+1) and 2^(2(k+1)+1):
5^(2(k+1)+1) + 2^(2(k+1)+1) = 5^(2k+3) + 2^(2k+3) = 5^3 * 5^(2k) + 2^3 * 2^(2k)
We can factor out 125 (which is divisible by 7) from the first term, and 8 (which is also divisible by 7) from the second term:
5^(2(k+1)+1) + 2^(2(k+1)+1) = 125 * 5^(2k) + 8 * 2^(2k)
We can rewrite 8 as 7+1:
5^(2(k+1)+1) + 2^(2(k+1)+1) = 125 * 5^(2k) + (7+1) * 2^(2k)
Distributing the 2^(2k) term and regrouping:
5^(2(k+1)+1) + 2^(2(k+1)+1) = 125 * 5^(2k) + 7 * 2^(2k) + 2^(2k)
Now we can use the inductive hypothesis that 5^(2k+1) + 2^(2k+1) is divisible by 7 to replace 5^(2k+1) + 2^(2k+1) with a multiple of 7:
5^(2(k+1)+1) + 2^(2(k+1)+1) = 125 * 5^(2k) + 7 * (5^(2k+1) + 2^(2k+1)) + 2^(2k)
By the inductive hypothesis, 5^(2k+1) + 2^(2k+1) is divisible by 7, so we can replace it with a multiple of 7:
5^(2(k+1)+1) + 2^(2(k+1)+1) = 125 * 5^(2k) + 7m + 2^(2k)
where m is some positive integer.
We can now see that 5^(2(k+1)+1) + 2^(2(k+1)+1) is divisible by 7, since it can be expressed as the sum of a multiple of 7 (i.e., 7m)