Problems & Puzzles: Puzzles

 Puzzle 395. (2^n+3^n+4^n+5^n+6^n)/2 – 10 Anton Vrba shares with us the following puzzle from his own: Let q(n)=(2^n+3^n+4^n+5^n+6^n)/2 – 10. Questions. 1. Can you explain/prove why q(3)=210 divides q(n) for any odd n > 1? 2. Can you explain why, for n even,  5 divides q(n) if n/2 is odd & 7 divides q(n) if n mod 6>0 3. Can you find a prime (or PRP) of the form q(n) or prove that none exist? (no PRP found for n < 10000).

Contributions came from Adam Stinchcombe, Farideh Firoozbakht, J. C. Rosa & J. K. Andersen:

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Question 1 of 395 is easy to resolve if you know the theory of linear recurrence relations. The key idea from that area is: x(n)=sum(ai*ri^n,i=1..k) generates the characteristic polynomial p(y)=product(y-ri,i=1..k) then when you replace y^i with x(n+i) you get the recurrence relation for x(n). Also, the converse, given a recurrence relation, you can solve for the ri and write formulas for x(n) [Binet forms].

For instance, write z(n)=q(2n+1)=4^n+(3/2)*9^n+2*16^n+(5/2)*25^n+3*36^n-10*1^n which leads to characteristic polynomial (y-1)(y-4)(y-9)(y-16)(y-25)(y-36) which leads to recurrence relation z(n+6)=91z(n+5)-3003z(n+4)+44473z(n+3)-296296z(n+2)+773136z(n+1)-518400z(n). Once you prove (just calculate!) that z(1),z(2),z(3),z(4),z(5),z(6) all have factors of 210, the recurrence proves that z(7) does also, and by induction, all z(n) do.

Similarly, part 2 can be proved using recurrence relations.  The divisible by 5 pattern is verified with the appropriate formula for z(n)=q(4n+2) with characteristic polynomial (x-1) (x-16) (x-81) (x-256) (x-625) (x-1296) and relation z(n+6)= 2275z(n+5)- 1516515z(n+4)+ 337967905z(n+4)- 22137475360z(n+3)+ 290539581696z(n+1)- 268738560000z(n).

Additionally, any residue pattern that repeats can be proved inductively.  You calculate that the initial pattern [0,0,0,0,0,3] leads to a repeat [0,0,0,0,0,3] and then that leads to a second repeat, etc., on forever.

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Farideh wrote:

q(n) = (1/2)*c(n) where c(n) = 2^n+3^n+4^n+5^n+6^n-20.

If n>1 is odd then 210 divides q(n).
Proof: Since n is odd we have the following four results.

1. c(n) = (-1)^n+0^n+1^n+(-1)^n+0^n-(-1)   (mod 3),  so c(n) = 0 (mod 3).
2. c(n) = 2^n+(-1)^n+0^n+1^n+(-2)^n-0   (mod 4),  so c(n) = 0 (mod 4).
3. c(n) = 2^n+(-2)^n+(-1)^n+0^n+(1)^n-0   (mod 5),  so c(n) = 0 (mod 5).
4. c(n) = 2^n+3^n+(-3)^n+(-2)^n+(-1)^n-(-1)   (mod 7),  so c(n) = 0 (mod 7)

Hence lcm(3, 4, 5, 7) = 420  divides c(n), so (1/2)*420 = 210 divides q(n).

(a). If n is even and n/2 is odd then 5 divides q(n).
(b). If n is even and mod(n, 6)>0 then 7 divides q(n).

Proof of (a): If n is even and n/2 is odd  then n is of the form 4k+2,
c(n) = 2^n+(-2)^n+(-1)^n+0^n+1^n (mod 5), so c(n) = 2^(n+1)+2 (mod 5)
or c(n)=2^(4k+3)+2 =16^k*8+2 (mod 5),
hence c(n) = 1*(-2)+2 = 0 (mod 5), so  5 divides q(n).

Proof of (b): Since n is even and mod(n, 6)>0 n is of the form 6k+2 or 6k+4.
1. If n = 6k+4 then c(n) = 2^n+3^n+(-3)^n+(-2)^n+(-1)^n-(-1) mod (7),
so c(n) = 2^(6k+5)+2*3^(6k+4) =(2^6)^k*32+2*(3^6)^k*81 (mod 7),
hence by FLT c(n) = 0 (mod 7),  so 7 divides q(n).

From Q1 & Q2 we deduce that if q(n) is prime or PRP then n is of the form 12*m.

Also we can show the following assertions.

1. If m = 1 (mod 3)  then 37 divides q(12m).
2. If m = 2 (mod 5)  then  61 divides q(12m).
3. If m = 4 (mod 11)  then 23 divides q(12m).
4. If m = 7 (mod 13)  then 79 divides q(12m).
5. If m = 4 (mod 17)  then 103 divides q(12m).
6. If m = 4 (mod 33)  then 397 divides q(12m).
7. If m = 3 (mod 35)  then 71 divides q(12m).
8. If m = 6 (mod 67)  then 269 divides q(12m).
9. If m = i (mod 130)  then 131 divides q(12m),
where i is in the set {4, 12, 69, 77} .
10. If m = i (mod 191)  then 383 divides q(12m),
where i is in the set {2, 86, 153} .

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Rosa wrote:

Question 1

We have : q(n)=(2^n+3^n+4^n+5^n+6^n)/2-10

fromwhere :  2q(n)=2^n+3^n+4^n+5^n+6^n-20

2q(n)=2^n(1+2^n)+3^n(1+2^n)+5^n-20

2q(n)=(1+2^n)(2^n+3^n)+5^n-20

if n is odd and >1 we have :

modulo 4: 2^n=0 ; 3^n=3 ; 5^n=1

2q(n)=(1+0)(0+3)+1-0=0

2q(n)=0 mod 4 implies q(n)=0 mod 2

modulo 3: 2^n=2 ; 3^n=0 ; 5^n=2

2q(n)=(1+2)(2+03)+2-(-1)=0

2q(n)=0 mod 3 implies q(n)=0 mod 3

modulo 5: 2^n=+/-2 ; 3^n=+/-3 ; 5^n=0

2q(n)=(1+/-2)(+/-5)+0-0=0

2q(n)=0 mod 5 implies q(n)=0 mod 5

modulo 7:  n is odd , n=2k+1

2^(2k+1)=2.4^k

3^(2k+1)=3.2^k

4^(2k+1)=4.2^k

5^(2k+1)=5.4^k

6^(2k+1)=6.1^k

2q(n)=2^n+3^n+4^n+5^n+6^n-20

2q(n)=2.4^k+3.2^k+4.2^k+5.4^k+6-6

2q(n)=(2+5)4^k+(3+4).2^k=0

2q(n)=0 mod 7 implies q(n)=0 mod 7

conclusion : if n is odd >1 then q(n) is divisible by 2,3,5,7 and therefore q(n) is divisible by 210

question 2 : n=4k+2  , we have  modulo 5:

2^(4k+2)=4

3^(4k+2)=4

4^(4k+2)=1

5^(4k+2)=0

6^(4k+2)=1

2q(n)=2^n+3^n+4^n+5^n+6^n-20

2q(n)=4+4+1+0+1-0=0

Thus if n is of the form 4k+2 then q(n) is divisible by 5

: n=2k  , we have  modulo 7

:                                    2^(2k)=4^k

3^(2k)=2^k

4^(2k)=2^k

5^(2k)=4^k

6^(2k)=1

2q(n)=2^n+3^n+4^n+5^n+6^n-20

2q(n)=4^k+2^k+2^k+4^k+1-(-1)

2q(n)=2(2^k+(2^k)^2+1)

q(n)=2^k+(2^k)^2+1  mod 7

If k=3x ,  we have : 2^k=2^(3x)=(2^3)^x=1  and q(n)=3 mod 7

If k=3x+1 , we have : 2^k=2  and q(n)=2+4+1= 0 mod 7

If k=3x+2 , we have : 2^k=4  and q(n)=4+4^2+1=21= 0 mod 7

Thus if n is  even of the form 6x+2  or 6x+4 then q(n) is divisible by 7, and if n=6x then q(n)=3 modulo 7.

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Question 3 : Two remarks:

If we want q(n) prime then n is necessarily of the form n=12k

1)   If k is of the form 3x+1 then q(n) is divisible par 37.

Indeed we have modulo 37 :

2^12=26

3^12=10

4^12=10

5^12=10

6^12=1

2^36=1  and  (2^36)^x=1   mod 37

2q(n)=2^(36x+12)+3^(36x+12)+4^(36x+12)+5^(36x+12)n+6^(36x+12)-20

2q(n)=2^12+3^12+4^12+5^12+6^12-20=26+10+10+10+1-20=37=0 mod 37

2q(n)=0 mod 37 implies q(n)=0 mod 37

Thus if q(n) is prime then n=36x or n=36x+24.

2)      If  k is of the form 5x+2 then q(n) is divisible by 61.

Indeed we have modulo 61 :

2^12=9

3^12=9

4^12=20

5^12=20

6^12=20

2^60=1  and  (2^60)^x=1   mod 61

2q(n)=2^(60x+24)+3^(60x+24)+4^(60x+24)+5^(60x+24)n+6^(60x+24)-20

2q(n)=2^24+3^24+4^24+5^24+6^24-20=81+81+400+400+400-20=1342=0 mod 61

2q(n)=0 mod 61 implies q(n)=0 mod  61

Thus if n=60x+24 then q(n) is not a prime , q(n)=0 mod 61.

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Andersen wrote:

1. and 2. 210 divides q(n) if 420 divides 2*q(n) = 2^n+3^n+4^n+5^n+6^n - 20.
For any fixed p and b: b^n (mod p) is periodic.
Then, for any fixed p, and k values b_1,...,b_k: b_1^n+...+b_k^n (mod p)
will also be periodic (and the period length will divide the least common
multiple of the k periods).
Brute force computation of the periodic values will easily show the wanted.
I haven't looked for a more elegant proof.

3. A search with PrimeForm/GW found no prime or prp for n < 150000.
The given observations show that if n is not divisible by 12, then 5 or 7
divides q(n). If n = 12m, then similar period observations will show that
q(n) has no prime factor below 23. Only some larger primes can ever divide
q(n), but some of them do it often - especially 37 which divides for n =
12*(3i+1). Some q(n) have no small prime factor, for example
q(60) = 289408609559171 * 84438651321055179349130110019677.
Based on these observations and the growth rate of q(12m), I guess there are
infinitely many primes but they are far apart.

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