BdMO Online Forum

(1) Show that if $a$and $b$ are relatively prime integers, then $gcd(a+b,a^{2}-ab+b^{2}) = 1$ or $3$.
(2) Show that if $a$ and $b$ are relatively prime integers, and $p$ is an odd prime, then $gcd(a+b,\frac{a^{p}+b^{p}} {a+b})=1$ or $p$.
(3)Show that if $a,m,n$ are positive with $m\neq n$,then $gcd(a^{2^{m}}+1,a^{2^{n}}+1)=1$ if $a$ is even and $2$ if $a$ is odd.

just apply $euclidean$ $algorithm$. nothing else.

Tahmid wrote:just apply $euclidean$ $algorithm$. nothing else.

I have solved $1$ and $2$ using $\text{Euclidean Algorithm}$.
$\text{Solution of 3}:$
WLOG,assume that $m>n$.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$.$\therefore gcd(a^{2^{m}}-1+2,a^{2^{n}}+1)=1$ if $a$ is even and $2$ if $a$ is odd.

tanmoy wrote:.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$.

how ?
$a=3$ ; $n=2$ ; $m=3$ breaks it .

Tahmid wrote:

tanmoy wrote:.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$.

how ?
$a=3$ ; $n=2$ ; $m=3$ breaks it .

I can't see $(a,n,m)=(3,2,3)$ yielding a contradiction.

In fact, if $A_n=a^{2^n}+1$ and $A_m=a^{2^m}-1$ and $m=n+j$ then \[a^{2^n}\equiv -1~\left(\bmod ~A_n\right)\Longrightarrow \left(a^{2^n}\right)^{2^j}\equiv (-1)^{2^j}\Longrightarrow a^{2^{n+j}}\equiv 1\Longrightarrow A_m\equiv 0~\left(\bmod~A_n\right).\]

Tahmid wrote:

tanmoy wrote:.$a^{2^{n}}+1$ is a divisor of $a^{2^{m}}-1$.

how ?
$a=3$ ; $n=2$ ; $m=3$ breaks it .

$3^{2^{2}}+1=3^{4}+1=82$ and $3^{2^{3}}-1=3^{8}-1=6560.6560\div 82=80$.

i thought $a^{2^{n}}=(a^{2})^{n}$ but you meant $a^{2^{n}}=a^{(2^{n})}$
sorry for the mistake .