341That's right.
To elaborate on what cele is saying, he is using that
1+x+x^2+x^3+x^4|1+x^m + x^{2m} + x^{3m} + x^{4m} when m is prime to 5.
Of course, the remaining part of the thread can be devoted to proving this result and a generalization.
Prove that for all n\ge1, 1+5^n+5^{2n}+5^{3n}+5^{4n} is composite
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17 Answers
341
9ans can be obtained by extending the argument i made in my second post
the path to follow had already been said by prophet sir
lets take
n= 5p.λ
now gvn
= (1+x+...x4) .(1 + x5 ........x5(λ-1)) /(1+...xλ-1 )
where x=5p 1≤ λ≤4
11I have proved for n as even, but for odd n .....I'm at a lossssss.......
11A prime p, (not equal to 5 or 7) satisfies the following property,
p^6 is congruent to 1(mod168), does that help in any way....
341celestine should be able to do this, because it was over a similar problem that he and I had our first encounter a long time ago (this should start off a treasure hunt :D)
62(1+x+x2+x3+x4)
x2{1/x2+1/x+1+x+x2}
x2 { (x+1/x)2 + (x+1/x) - 1 }
This was all that I could think :(
but that does not seem to help...
341I just realized that by modifying the problem, it becomes somewhat more accessible.
Prove that if n is of the form 5pq with gcd(5,q) = 1then 1+5^n+5^{2n} + 5^{3n} + 5^{4n} is not a prime.
1OK so this is not an easier one... I am sorry for my mistake.. I thought this is a GP and some manipulations may lead to the answer..
341I have seen the solution. It involves some convoluted factorising.
62that was the first thing that came to my mind.. but somehow
1+5+52 =131 is prime! So i din give this hint!
9still thinking
gvn = 55n-1 /5n-1 = 781 ( 1+...55(n-1))/(1+......5n) = 781Xp/q
where p and q are coprime so 781/q = integer necessarily
when n≠5λ
q cant be 781 ( see why ?)
781p/q = rp with r≠1 so its composite
for n=5 i donno how
341to illustrate how helpful that comment is:
consider 1 + 2 + 22+23+24 . This is a GP