39no single taker?
1) Prove that there exist arbitrarily long arithmetic progressions with all the terms being powers of integers.
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5 Answers
9ofc dude its obvious
eg 1 1 1 1 1 1
4 4 4 4 4 4 4 4 4 4 ..........
62even i was going to give the same solution :D
but then i thought that it wud be unfair to the question ;)
I am sure the question has the phrase "unequal terms"
I have seen the question but dont know the prooff.. the proof i saw had arguments on Real analysis!
9i guess proof gots to deal with the fact we got infinite no of powers of integers
39By rk denote member of the finite arithmetic progression
n!+1+k(n-1)!,k=1..n.
It is obvious gcd(ri,rj)=1.
Let ai=si*r1*r2*...*rn/ri,
si is positive integer and satisfies the condition
ri|(ai+1) (gcd(r1*r2*...*rn/ri ,ri)=1
==>si exists.
Denote m=r1a1*...*rnan.
The sequence r1*m, r2*m,..., rn*m is required since rk*m=b^rk for some positive integer b.