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}


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

  • $\begingroup$ How do you get to $v_2(b+1)=1$? Can you please explain this step a bit more? $\endgroup$ – thinkingeye Apr 19 '20 at 17:07
  • $\begingroup$ $b+1\equiv_4 2$, so $2|b+1$ and $4\not |b+1$ $\endgroup$ – L3435 Apr 19 '20 at 17:09
  • $\begingroup$ 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 $v_2(a+1)=v_2(b-1)$ $\endgroup$ – L3435 Apr 19 '20 at 22:15
  • $\begingroup$ 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? $\endgroup$ – thinkingeye Apr 19 '20 at 22:25
  • $\begingroup$ Otherwise $ab>(2^n-1)(2^n+1)=ab$, a cobtradiction. $\endgroup$ – L3435 Apr 19 '20 at 22:46

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