Problems & Puzzles:
Puzzles
Puzzle 153.
Espinosa's puzzle
José Espinosa, from Santiago de Chile, posed the
following problem
 Let p be a prime number greater than 3 and q a
prime number.
 Let g(n)=2^(2+n(p1))+3^(3+n(p1))+5^(5+n(p1))+7^(7+n(p1))+...+
q^(q+n(p1))
 Let f(n)=48g(n)(n5)(n7)g(13)+4(n5)(n13)g(7)3(n7)(n13)g(5)
Prove that for every integer
nonnegative n: f(n) is divisible by p^3.
He also sent to me a hint
that I will provide to those of you who need it.
But maybe can help enough to know that Mr.
Espinosa currently maintains a math
site
devoted to the mathematical induction.
Jan van Delden wrote:
Focus on the effect f has on one term of g:
Consider one term t(n,r)=r^(r+n*(p1))
And h(n,r):=48*t(n,r)(n5)(n7)t(13,r)+4(n5)(n13)t(7,r)3(n7)(n13)t(5,r)
We then have g(n)= t(n,2)+t(n,3)+t(n,5)..+t(n,q), and f(n)=h(n,2)+h(n,3)+h(n,5)+..h(n,q).
If we can prove that for every prime r: h(n,r)=0 mod p^3 we're done.
As suggested we use induction on n.
n=0:
h(0,r)=r^r [48(n5)(n7)
r^(13(p1))+4(n5)(n13)r^(7(p1))3(n7)(n13)r^(5(p1))]
r^(p1)=1 mod p if (r,p)=1 i.e. we can write r^(p1)=Ap+1, for some A
integer if (r,p)=1.
h(0,r)=r^r
[48(n5)(n7)(Ap+1)^13+4(n5)(n13)(Ap+1)^73(n7)(n13)(Ap+1)^5]
It is now straightforward to show (use binomal of Newton, a bit awkward)
that the terms between the brackets corresponding to p^0,p^1 and p^2 are
0, so
h(0,r)=0 mod p^3.
If (r,p)>1 and because r,p are prime we have r=p. We then have:
h(0,r) = h(0,p)= p^p [48 + extra terms each at least=0 mod p^(5(p1)].
So h(0,r)=0 mod p^3, if p>=3.
[Actually 48=2^4*3. With p=3: p^(5(p1))=0 mod p^10 and with p=2:
p^(5(p1))=0 mod p^5, so h(0,3)=0 mod 3^4 and h(0,2)=0 mod 2^6]
n+1: Induction step
h(n+1,r)h(n,r) = 48 [t(n+1,r)t(n,r)]+(112n)t(13,r)+4(2n17)t(7,r)+3(192n)t(5,r)
= r^r {48 r^(n*(p1)) [r^(p1)1] + (112n)
r^(13(p1))+4(2n17)r^(7(p1))+3(192n)r^(5(p1))}
If (r,p)=1, we proceed as before and find that h(n+1,r)h(n,r)=0 mod
p^3.
If (r,p)=p we see that the term p (or r if you will) is even more
abundant as a factor compared with the h(0,r) case! [Also for p=2 or
p=3]
Assuming h(n,r)=0 mod p^3, using the above we have h(n+1,r)=0 mod p^3.
Hence
h(n,r)=0 mod p^3 with p>=2.
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