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To provide full context of the practice question I'm attempting, it is as follows:

For every integer n such that n ≥ 8, there exist nonnegative integers an and bn such that 3an + 5bn = n.

Write a proof of the claim using the strong form of mathematical induction with the integer 10 as breakpoint — such that that n = 8, n = 9 and n = 10 would all be considered in the basis.

This is my attempt at the proof, but I can't seem to get past applying the I.H. in order to reach my goal in the inductive step.

Basis
If n = 8, then 3a8 + 5b8 = 8
To satisfy the above, set a8 = 1 and b8 = 1, therefore 3(1) + 5(1) = 3 + 5 = 8

If n = 9, then 3a9 + 5b9 = 9
Set a9 = 3 and b9 = 0, therefore 3(3) + 5(0) = 9 + 0 = 9

If n = 10, then 3a10 + 5b10 = 10

Set a10 = 0 and b10 = 2, therefore 3(0) + 5(2) = 0 + 10 = 10

Inductive Hypothesis
Let k be an integer such that k ≥ 10. It is necessary and sufficient to use the following:

  • Inductive Hypothesis: 3an + 5bn = n for every integer n such that 8 ≤ n ≤ k.

Inductive Claim
3ak+1 + 5bk+1 = k+1

Since for every integer n, 8 ≤ n ≤ k, there are ak-2 and bk-2, then it follows that:

3ak-2 + 5bk-2 = k - 2 (Starting Point)

..

..

..

3ak+1 + 5bk+1 = k+1 (Goal)

Now I know where to begin and the goal I need to reach for my inductive step but I am unsure as to how to reach my goal.

Any guidance would be much appreciated!

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3 + 5 = 8, 3×3 = 9, 5x2 = 10

The next step is 11 = 8 + 3.
Since 8 has been answered, an answer for 11 is apparent.
Likewise for n > 10, by induction hypothesis,
there is a solution for n - 3, so a solution for n.

Exercise. Show every integer >= 8 is a multiple of 3
or 5 + a multiple of 3 or 10 + a multiple of 3
as an immediate result of the above proof.

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