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 $F_4$.
IMO LONGLISTED PROBLEM 1976
Re: IMO LONGLISTED PROBLEM 1976
Notice that $19^{1976}+76^{1976}$ has the form $a^4+4b^4$
(The name is Fermat)
(The name is Fermat)
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Nur Muhammad Shafiullah | Mahi
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Nur Muhammad Shafiullah | Mahi
- Tahmid Hasan
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Re: IMO LONGLISTED PROBLEM 1976
working with sophie germain identity and congruences will do great
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Re: IMO LONGLISTED PROBLEM 1976
$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}+a^{2n-1}b+...+b^{2n})$ will be very helpful identity to solve the problem.
Re: IMO LONGLISTED PROBLEM 1976
$19^{1976}+76^{1976}=19^{976}(1+4^{1976})$ and $1+4^{1976}=1+(2^{16})^{247}$ which is divisible by $2^{2^4}+1=F_4$.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 $F_4$.
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 $0$.