IMO LONGLISTED PROBLEM 1976

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MATHPRITOM
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IMO LONGLISTED PROBLEM 1976

Unread post by MATHPRITOM » Mon May 02, 2011 7:31 pm

Prove that the number $19^{1976}+76^{1976}$:
a)is divisible by the (Fermet) prime number $F_4$=$2^{2^4}+1$.
b)is divisible by at least four distinct primes other than $F_4$.

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*Mahi*
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Re: IMO LONGLISTED PROBLEM 1976

Unread post by *Mahi* » Wed May 04, 2011 5:13 pm

Notice that $19^{1976}+76^{1976}$ has the form $a^4+4b^4$
(The name is Fermat)
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Tahmid Hasan
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Re: IMO LONGLISTED PROBLEM 1976

Unread post by Tahmid Hasan » Thu May 05, 2011 6:15 pm

working with sophie germain identity and congruences will do great
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MATHPRITOM
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Re: IMO LONGLISTED PROBLEM 1976

Unread post by MATHPRITOM » Fri May 06, 2011 9:32 pm

$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}+a^{2n-1}b+...+b^{2n})$ will be very helpful identity to solve the problem.

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Masum
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Re: IMO LONGLISTED PROBLEM 1976

Unread post by Masum » Mon May 09, 2011 2:39 pm

MATHPRITOM wrote:Prove that the number $19^{1976}+76^{1976}$:
a)is divisible by the (Fermat) prime number $F_4$=$2^{2^4}+1$.
b)is divisible by at least four distinct primes other than $F_4$.
$19^{1976}+76^{1976}=19^{976}(1+4^{1976})$ and $1+4^{1976}=1+(2^{16})^{247}$ which is divisible by $2^{2^4}+1=F_4$.
We are sure that $19$ divides this.
Now, note that $247=13\cdot 19$
So, it is divisible by $1+2^{208}$ and $1+2^{304}$. They are co-prime t o each other. Now complete this.
One one thing is neutral in the universe, that is $0$.

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