# 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)?

\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}

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 $$c=0\not\in\mathbb{N}.$$

• How do you get to $v_2(b+1)=1$? Can you please explain this step a bit more? – thinkingeye Apr 19 '20 at 17:07
• $b+1\equiv_4 2$, so $2|b+1$ and $4\not |b+1$ – L3435 Apr 19 '20 at 17:09
• If $v_2(x)\ne v_2(y)$, then $v_2(x+y)=\min\{v_2(x), v_2(y)\}$, because $p\cdot 2^a+q\cdot 2^b=2^{b}(2^{a-b}p+q)$. Seeing as $v_2((a+1)+(b-1))\ne\min\{v_2(a+1), v_2(b-1)\}$, we have $v_2(a+1)=v_2(b-1)$ – L3435 Apr 19 '20 at 22:15
• Thanks. I got it. The only thing left is: Ok, $x = y = 1$ is obviously a solution. I got to a similar result before during my attempts. But, how do I know that this is the only solution? – thinkingeye Apr 19 '20 at 22:25
• Otherwise $ab>(2^n-1)(2^n+1)=ab$, a cobtradiction. – L3435 Apr 19 '20 at 22:46