<s>[INST] How to prove that $a + b \neq 2^{n+1} (2c+1) $ with $ab = 4^n - 1$ and $a,b, c, n \in \mathbb{N}$? How to prove that $a + b \neq 2^{n+1} (2c+1) \quad \text{; with } ab = 4^n - 1 \text{ and } a,b, c, n \in \mathbb{N}$ (without zero)?
I already know that:
\begin{align}
a,b &\equiv 1 \pmod{2} \\
a + b &\equiv 0\pmod{4}
\end{align}
And because of the symmetry, I can define:
\begin{align}
a &\equiv 1 \pmod{4} \\
b &\equiv 3 \pmod{4}
\end{align}
For even $n$, I am already able to prove the statement by:
\begin{align}
2^{n+1} (2c+1) &\equiv 2 \pmod{4} \\
2^{n+1} (2c+1) &\not\equiv a + b \pmod{4}
\end{align}
For odd $n$, however, I am stuck. From the perspective of the congruence, the equation could hold for odd $n$ by $n+1 = 2m; \text{ with } m \in \mathbb{N}$ then:
\begin{align}
4^m (2c+1) &\equiv 0 \pmod{4} \\
4^m (2c+1) &\equiv a + b \pmod{4}
\end{align}
 [/INST] Define $v_2(n)$ to be the exponent of the highest power of $2$ that divides $n$, for example $v_2(24)=v_2(2^3\cdot 3)=3$ and $v_2(7)=0$.
Suppose that $a+b=2^{n+1}(2c+1)$. This gives $$v_2(a+b)=n+1.$$ It is not hard to see that $$n+1=v_2(4^n+a+b)=v_2((a+1)(b+1))=v_2(a+1)+v_2(b+1).$$ As seen above, WLOG $b\equiv_4 1$, so $v_2(b+1)=1$. This gives $v_2(a+1)=n,$ but because $v_2((a+1)+(b-1))=n+1$, we must also have $v_2(b-1)=n$. Thus, we have $a=2^n\cdot x-1$ and $b=2^n\cdot y+1$ for some odd $x$ and $y$. Plugging this in $ab=4^n-1$ we get $x=y=1$, so $c=0\not\in\mathbb{N}.$
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