Mistakes Give Rise to Problems

Clearly if $a=0$ then $a^{m+n} = a^m + a^n$ is true for all $\left(m,n\right)$ Which means that $a=0$ yields $10*10 = 100$ solutions.

If $a \neq 0$ we get:

$a^ma^n = a^m + a^n$

$a^m\left(a^n-1\right) = a^n$

$a^n - 1 = a^{n-m} \Rightarrow 1 = a^n - a^{n-m}$

This implies that for some non-zero integer $a$ , there are two integer powers of $a$ with a difference of $1$

Clearly this is only true if $a^{n-m} = 1 \Rightarrow a^n = 2 \Rightarrow a =2 \wedge n=m=1$

This then means that there are $100+1 = 101$ possible triplets.

$a = 0$ always work for any $m,n \neq 0$ (100 ways)

For $a \neq 0$ , let $a^{m} = x, a^{n} = y$ such that $x,y \neq 0$

The equation becomes $xy = x+y$ .

Factor as $(x-1)(y-1) = 1$

Which gives $(x,y) = (0,0), (2,2)$ .

$a^{m} = 2, a^{n} = 2$

Which gives $(a,m,n) = (2,1,1)$ . (1 way)

Therefore, the number of ways $= \boxed{101}$ .

$a^{m+n} = a^m + a^n$ If $a = 0$ , we get $0^{m+n} = 0^m + 0^n$ From here, we can see that any value of $m$ and $n$ will satisfy the equation above. Therefore, all triplets in the form of ( $0,m,n$ ) satisfy the equation above. There are $10$ possible values of $m$ and $n$ each, so we have $100$ different triplets in the form of ( $0,m,n$ ).

If $a \neq 0$ , $a^{m+n} = a^m + a^n$ $a^{m+n} - a^m - a^n = 0$ $a^ma^n - a^m + 1 - a^n = 1$ $a^m(a^n - 1) - 1(a^n - 1) = 1$ $(a^m - 1)(a^n - 1) = 1$ The question has stated that the triplets are integers. That means that $a^m$ and $a^n$ are both integers, which also means that $(a^m - 1)$ and $(a^n - 1)$ are both integers.

Now, the only possible way for two integers to have a product of 1 is to have both integers be 1, that is $1 * 1 = 1$

From there, we can infer that $(a^m - 1) = (a^n - 1) = 1$ $a^m = a^n = 2$

If $a \geq 3$ , no integer value of $m$ or $n$ can make $a^m = a^n = 2$

If $a = 1$ , $a^m = a^n = 1$ for all values of $m$ and $n$

Therefore, the only possible value of $a$ that isn't $0$ is $2$ .

When $a = 2$ $2^m = 2^n = 2$ It is pretty obvious that $m = n = 1$

Therefore, the only triplet with $a \neq 0$ that satisfies the equation is $(2,1,1)$

Finally, the total number of solutions is $1 + 100 = \boxed{101}$ solutions.

$a^{m+n} = a^m + a^n$

When $n \neq 0$ and $m \neq n$ ,

$a^m = a^{m-n} + 1$

$a^m - 1 = a^{m-n}$

This is invalid when $m<n$

When $m>n$ there are no solutions because

If $a$ is odd, LHS = even and RHS = odd

If $a$ is even, LHS = odd and RHS = even

When $m=n$

$a^m -1 = 1$ ; There is a solution of $(2, 1, 1)$

When $a = 0$

There are solutions for all $m$ and $n$ ; 100 triples.

Total = $\boxed{101}$ triples

First, the case when a=0 works for all cases because 0^x=0 and 0+0=0. Also, since there are 10 possible values for m,n, there are 100 possible triples of (0,m,n).

Next for a>0, we divide both sides by a^n and get $a^m=a^{m-n}+1$ . This means that the difference between $a^m$ and $a^{m-n}$ must be 1. However, this is only achievable if a=2 and m=n=1. The proof that this doesn't work is that there is no positive integer that when raised to a power is 1 greater than itself except for 2. Finally, we add the 100 cases when a=0 and the special case when a=2 and our total count is 101.

Side note: if an integer x>2 and integers p>n>0, $x^p\neq x^{p-n}+1$

First we know that it will work with any combination with zero as a base so we have $10\times10=100$ solutions first.

Now, if we rephrase the equation:

$a^{m+n}=a^m+a^n$

$\implies a^m-1=a^{m-n}$

We can observe that the right side ( $a^{m-n}$ ) is a multiple of $a$ and the right side is a multiple of $a$ decreased by one. Obviously, these are not equal, except for one case:

In $a=2$ , $m=1$ , $m=2$

$2^1-1=2^{1-1}$

$\implies 1=2^0$ , Which is true.

$\therefore$ We have $\boxed{100+1=101}$ solutions.