Method 1: By Mathematical Induction
(1) We want to prove for all odd integers n > 0 the statement P(n) that 9ⁿ + 1ⁿ is a multiple of 10.

P(1) is true. If P(k) is true then consider P(k+2):

9^k+2 + 1^k+2 = 81(9^k + 1^k)-80

so P(k+2) is true and the result is true for all odd numbers by the axiom of induction.

Similarly 7¹ + 3¹ is a multiple of 10 and if 7^k + 3^k is a multiple of 10 then

7^k+2 + 3^k+2 = 49(7^k) + 9(3^k) = 40(7^k) + 9(7^k + 3^k)

so this is also a multiple of 10 and the result is true for all odd numbers by the axiom of induction. Note that 9²+1² and 7²+3² are not multiples of 10 so the result is not true for even numbers.

(2) We have 8⁰-2⁰ and 8²-2² are both multiples of 10. If 8^k-2^k is a multiple of 10 then

8^k+2 - 2^k+2 = 64(8^k) - 4(2^k) = 60(8^k) +4(8^k-2^k)

so this is a multiple of 10 and the result is true for all even numbers. Similarly 6²-4² is a multiple of 10 and if 6^k-4^k is a multiple of 10 then

6^k+2 - 4^k+2 = 36(6^k) - 16(4^k) = 20(6^k) + 16(6^k-4^k)

which is a multiple of 10 and so the result is true for all even numbers.

Note that 8 - 2 and 6 - 4 are not multiples of 10 and the result is not true for odd powers.

(3) The corresponding results are 9ⁿ - 1ⁿ and 7ⁿ - 3ⁿ are multiples of 10 when n is an even number which can be proved by induction.

(4) Similarly 8ⁿ + 2ⁿ and 6ⁿ + 4ⁿ are multiples of 10 when n is an odd number which can be proved by induction.

Proof using the Binomial Theorem

7ⁿ + 3ⁿ = 7ⁿ + (10-7)ⁿ = 7ⁿ + 10ⁿ + ..... + (-7)ⁿ.

All the terms of the Binomial expansion involve powers of 10 apart from the term (-7)ⁿ so the expression is a multiple of 10 when n is odd but not when n is even. The same method works for all other parts of the question.