for prime (a,b).....
find prime (a,b) sothat $a^b+b^a$ is also a prime.
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- afif mansib ch
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Re: for prime (a,b).....
(1,2),(2,3)
- nafistiham
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Re: for prime (a,b).....
As far as I know $1$ is not prime.afif mansib ch wrote:(1,2),(2,3)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
- afif mansib ch
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Re: for prime (a,b).....
ok then (2,3)only!!!
Re: for prime (a,b).....
why don't you post your proof ? giving just the answers reminds me about talking answers in exam hall.afif mansib ch wrote:ok then (2,3)only!!!
Try not to become a man of success but rather to become a man of value.-Albert Einstein
- afif mansib ch
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Re: for prime (a,b).....
1st i tried with trivial magnitudes like 1,2,3.
then for any prime a and b>3
\[a^b,b^a\equiv 1(mod2)\]
so\[a^b+b^a=p\equiv 0(mod2)\]
which can't be prime.
then for any prime a and b>3
\[a^b,b^a\equiv 1(mod2)\]
so\[a^b+b^a=p\equiv 0(mod2)\]
which can't be prime.
Re: for prime (a,b).....
$1$ is neither prime nor composite.afif mansib ch wrote:(1,2),(2,3)
One one thing is neutral in the universe, that is $0$.
- Tahmid Hasan
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Re: for prime (a,b).....
what about considering the case when one is $2$afif mansib ch wrote:1st i tried with trivial magnitudes like 1,2,3.
then for any prime a and b>3
\[a^b,b^a\equiv 1(mod2)\]
so\[a^b+b^a=p\equiv 0(mod2)\]
which can't be prime.
বড় ভালবাসি তোমায়,মা
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Re: for prime (a,b).....
Just take mod (3)
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
- afif mansib ch
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Re: for prime (a,b).....
as sourav via said i proved it for 2 also.(but only 15 days left for s.s.c.so i enter the forum hiding.it's hard to post full sol bcause 'f time)lol.