Convincing Myself of Stamp Induction Induction Proof? 
Use mathematical induction (and proof by division into cases) to show that any postage of at least 12 cents can be obtained using 3 cent and 7 cent stamps.

So for this I understand that it can be solved using induction without a strong hypothesis.
Base cases:
$$ n = 12 : 3*4 + 7*0$$
$$ n = 13 : 3*2 + 7*1$$
$$ n = 14 : 3*0 + 7*2$$
Induction hypothesis:
Assume for some $k>14$ that $k = 3a + 7b$ for $a, b \in Z$.
Induction step:
We will show that $k + 1$ can be made up of 3 and 7 cent stamps.
$$k+1 = 3a + 7b + 1$$
$$k+1 = 3a + 7b + 7 -6$$
$$k+1 = 3a - 6 + 7b + 7$$
$$k+1 = 3(a-2) + 7(b+1)$$
So when $a > 2$ we can make $k + 1$ stamps.
Is it at this point that I need to also clarify when $a = 1$ and $a = 0$ that other cases of $k$ will hold as well? Have I already proven my original statement? Or is this covered because I have shown the pattern in the base cases?
 A: Simplest version: [Induction by $n+3$]
It holds for $n=12,13,14$ as you already showed.
If it holds for $n$, then
$$n=3a+7b$$
$$n+3=3(a+1)+7b$$
Therefore it also holds for $n+3$.
So, it holds for all $n\geq12$.
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Different version: [Induction by $n+1$]
Assume it holds for $n$, and $a\geq2,b\geq1\text{ or }a\geq0,b\geq2\text{ or }a\geq4,b\geq0$ holds, then
$$n=3a+7b\quad (a\geq2,b\geq1\text{ or }a\geq0,b\geq2\text{ or }a\geq4,b\geq0)$$
i) If $a\geq2,b\geq1$,
$$n+1=3(a-2)+7(b+1)\quad (a-2\geq0,b+1\geq2)$$
ii) If $a\geq0,b\geq2$,
$$n+1=3(a+5)+7(b-2)\quad (a+5\geq4,b-2\geq0)$$
iii) If $a\geq4,b\geq0$,
$$n+1=3(a-2)+7(b+1)\quad (a-2\geq2,b+1\geq1)$$
Therefore it also holds for $n+1$ and still $a\geq2,b\geq1\text{ or }a\geq0,b\geq2\text{ or }a\geq4,b\geq0$ holds.
So, it holds for all $n\geq12$.
A: Yes you do need to cover the cases of when $a=1$ or $0$.
This does not even cover the case of $n=15$
If you trace the logic, it says:
$$14=3*0+7*2$$
Thus,
$$15 = 3*(-2)+7*3$$
But this is invalid because you can't have $-2$ stamps.
A: Base case: $12=3\cdot 4+7\cdot 0$, $13=3\cdot 2+7\cdot 1$, $14=3\cdot 0+7\cdot 2$.
Inductive hypothesis: Assume that for some (not for all, like you said) $k\ge 12$ we have $k=3m_1+7n_1,k+1=3m_2+7n_2,k+2=3m_3+7n_3$ for some $m_i,n_i\ge 0$.
Inductive step: $k+3=3(m_1+1)+7n_1$ and $k+4=3(m_2+1)+7n_2$ and $k+5=3(m_3+1)+7n_3$.
It also follows from the Chicken McNugget Theorem:

If $a,b\ge 1, \gcd(a,b)=1$, then the greatest integer not of the form $am+bn$ for some $m,n\ge 0$ is $ab-a-b$.

A: The base of induction is good: you can start from $n=14$. 
Now beware: the induction hypothesis is that for $k\ge14$ (not $k>14$), $k=3a+7b$ for non negative integers $a$ and $b$.
At this point you can divide into cases for $k+1=3a+7b+1$.


*

*If $a=0$ or $a=1$, then $b\ge2$ (because $3+7=10<k$) and $$3a+7b+1=3a+7(b-2)+14+1=3(a+5)+7(b-2);$$

*If $a\ge2$, then $3a+7b+1=3(a-2)+7(b+1)$.

