1) Test n = 1
7 + 11
= 18
= 9 x 2 hence divisible by 9
.: True for n = 1
Assume statement is true for n = k (where k is an odd integer)
i.e. 7<sup>k</sup> + 11<sup>k</sup> = 9M (where M is an integer)
Need to prove the statement is true for n = k + 2 (since k is an odd integer, the next odd integer is k + 2)
i.e 7<sup>k + 2</sup> + 11<sup>k + 2</sup> = 9N (where N is an integer)
LHS = 7<sup>k + 2</sup> + 11<sup>k + 2</sup>
= 7<sup>k</sup>.7² + 11<sup>k + 2</sup>
From assumption: 7<sup>k</sup> = 9M - 11<sup>k</sup>
= 7²(9M - 11<sup>k</sup>) + 11<sup>k</sup>.11²
= 49.9M - 49.11<sup>k</sup> + 121.11<sup>k</sup>
= 49.9M + 72.11<sup>k</sup>
= 9(49M + 8.11<sup>k</sup>)
= 9N (where N = 49M + 8.11<sup>k</sup>)
= RHS
If it is true for n = k, then it is also true for n = k + 2. Since it is true it is true for n = 1, it follows by induction that it is true for all odd integers n ≥1.
2) Test n = 1
7 + 3
= 10 x 1 hence divisible by 10
.: True for n = 1
Assume statement is true for n = k (where k is an odd integer)
i.e. 7<sup>k</sup> + 3<sup>k</sup> = 10P (where P is an integer)
Need to prove the statement is true for n = k + 2 (since k is an odd integer, the next odd integer is k + 2)
i.e 7<sup>k + 2</sup> + 3<sup>k + 2</sup> = 10Q (where Q is an integer)
LHS = 7<sup>k + 2</sup> + 3<sup>k + 2</sup>
= 7<sup>k</sup>.7² + 3<sup>k + 2</sup>
From assumption 7<sup>k</sup> = 10P - 3<sup>k</sup>
= 7².(10P - 3<sup>k</sup>) + 3<sup>k</sup>.3²
= 49.10P - 49.3<sup>k</sup> + 9.3<sup>k</sup>
= 49.10P - 40.3<sup>k</sup>
= 10(49P - 4.3<sup>k</sup>)
= 10Q (where Q = 49P - 4.3<sup>k</sup>)
= RHS
If it is true for n = k, then it is also true for n = k + 2. Since it is true it is true for n = 1, it follows by induction that it is true for all odd integers n ≥1.