To find the integer solutions of an indeterminate equation and prove that there are no other solutions, where all variables are positive integers and n can be regarded as a constant, let's first analyze the given situation where by observation we found that k=1,n1=n2=n is a solution to the equation. However, the question remains: Is this the only possible solution?
$$k2^{2n}-(2k-1)2^n+k-1=k2^{n_1+n_2}-2^{n_2}$$
The only solution in positive integers to equation $$2^{2n} k−(2k−1)2^n+k−1=2^{a+b} k−2^{a}$$ is $k=1$, $a=b=n$. To prove this claim, we use the elementary inequality $2^b -1\ge 2^{b-1}$ and remark that in any solution in positive integers one necessarily has $k$ odd.
Assume there is a solution with $k\ge 3$. The left-hand side is less than $2^{2n} k$, while the right-hand side is greater than $2^a (2^b-1)k \ge 2^{a+b-1}k$. Hence, $$2n \ge a+b.$$
To get a lower bound for $a+b$, rewrite the equation as $2^n+2^{a}−1=\bigl(2^{a+b} -(2^n-1)^2 \bigr)k$. The LHS being positive, so is the RHS. Therefore, $$ a+b> 2n-2.$$
There are two cases to examine.
Case $a+b=2n$. The equation becomes $2^n+2^a-1=(2^{n+1}-1)k$. The RHS is $\ge 3(2^{n+1}-1) > 2^{n+2}$, so we conclude that it holds $a\ge n+2$. This in turn implies $k=2^n c+1 $ for some odd positive integer $c$. Writing the equation in terms of $c$, one gets $2^{a-n}-1= (2^{n+1}-1) c$. Here, the LHS is less than $2^{n-1}$ (remember $2n=a+b\ge a+1$), while the RHS is at least $2^{n+1}-1 >2^n$.
Case $a+b=2n-1$. Note that from $a$, $b\ge 1$ it follows that $n\ge 2$. The equation is now $\bigl( 2^{2n-1}k-(2k-1)2^n\bigr) +(k-1) +2^a=0$. As both differences within parantheses are positive for $n$, $k>1$, this equality is false.
