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Find gcd($2^{19} + 1$; $2^{86} + 1$)

It would be easy to give a formal proof for any gcd($2^{n} + 1$; $2^{m} + 1$) based on Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$ if $m$, $n$ were uneven but the problem is: $86$ is an even number. What to do then? Can i solve it without finding an ultimate solution for any n, m? (Like, an easier way for this exact problem)

mhmhhmhmhm
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    The problem with that solution is not that $86$ is even, it's that it's not $-1$, but $+1$. – Rushabh Mehta Oct 17 '19 at 20:58
  • Well, i could follow the same logic? The difference would be that $2^m ≡ $ -1(mod d) and $2^n ≡ $ -1(mod d). But when we try to prove the fact that $2^p + 1 | 2^m + 1$ if $p | m$ we figure out that it is true only when $m$ is uneven and the same for $n$ (because of the abbreviation formula) – mhmhhmhmhm Oct 17 '19 at 21:05

3 Answers3

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So both numbers are odd, so there is no common factor $2$.

Any common factor divides the difference $2^{86}-2^{19}=2^{19}\left(2^{67}-1\right)$ and hence (by the earlier remark) $2^{67}-1$

Now any common factor of $2^{19}+1$ and $2^{67}-1$ divides their sum and hence $2^{67}+2^{19}=2^{19}\left(2^{48}+1\right)$ and hence $2^{48}+1$.

Note that this shows you can reduce $86$ by $2\times 19$.

In fact you can also exploit the fact that $2^{19}+1$ is a factor of $2^{38}-1$ and $2^{86}+1$ is a factor of $2^{172}-1$ and you can compute the common factor of these larger numbers with what you already know.

I reckon that is enough of a clue to get you started (there are other things you might notice too).

Mark Bennet
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Let $d$ be the gcd. Note: $$d\mid 2^{67}(2^{19}+1)-(2^{86}+1)=2^{67}-1\\ d\mid 2^{48}(2^{19}+1)-(2^{67}-1)=2^{48}+1\\ d\mid 2^{29}(2^{19}+1)-(2^{48}+1)=2^{29}-1\\ d\mid 2^{10}(2^{19}+1)-(2^{29}-1)=2^{10}+1\\ d\mid 2^{9}(2^{10}+1)-(2^{19}+1)=2^{9}-1\\ d\mid (2^{10}+1)-2^{1}(2^{9}-1)=3\\ d\in \{1,3\}\\ 2^{86}+1\equiv (-1)^{86}+1\equiv 2 \pmod{3} \Rightarrow 3\not\mid 2^{86}+1$$ Hence, $\gcd(2^{19}+1,2^{86}+1)=1$.

farruhota
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Note that by FLT, and the fact that $\mathbb Z/p\mathbb Z$ is a field, we have that for any odd prime $p$ and any $k\in\mathbb N$,

$$p\mid2^{\frac{(2k-1)\cdot(p-1)}2}+1$$

So, for both $19$ and $86$, we are looking for odd primes $p$ such that there exists a $k\in\mathbb N$ with $$2x=(2k-1)\cdot(p-1)$$($x$ is the exponent here).

For $2x=38$, $p-1$ has to be $38$ or $2$. However, $p=39$ isn't prime, so we are left with $p=3$. So, we know that $3|2^{19}+1$. In fact, this tells us that $3$ is the only prime less than $\sqrt{2^{19}+1}$ that is a factor.

This makes the remainder of the question pretty easy. Can you take it from here? Hint: 3 isn't a factor of $2^{86}+1$, so there are only two possible answers left.

Rushabh Mehta
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