Problems & Puzzles: Puzzles

Puzzle 431. p = (q/r)3 + (s/t)3 Patrick Capelle sent the following puzzle: We are interested by prime numbers p such that p = (q/r) 3  + (s/t) 3 , where q,r,s,t are prime numbers and q/r is different from s/t. Examples : 19 = (5/2)3 + (3/2) 3 13 = (2/3) 3 + (7/3) 3 Here, in each case, the denominator is the same for the two fractions. So, we can write : 19 = (53 + 33)/23 13 = (23 + 73)/33 They are solution of the Generalized Fermat Equation xn + yn = c.zn  for n = 3.  Questions : 1. Can you give other solutions for p ? 2. What's the smallest p? Is it 13? Can you justify your answer? 3. Is there a prime number p when q, r, s, t are distinct?

---

Contributions came from J. Wroblewski & Farideh Firoozbakht.

***

J. Wroblewski wrote:

Q3. We MUST have r=t for (q/r)^3+(s/t)^3 to be an integer.

Q1,2. We get the equation

p*r^3=q^3+s^3=(q+s)*(q^2-qs+s^2),

where we can assume q<s.

If r=2 we have q+s=8, leaving q=3, s=5.

p=2 is ruled out by Euler. [Go to: http://cboyer.club.fr/multimagie//English/SquaresOfCubes.htm
Search for word "Euler". That is all I know - apparently Euler showed that the sum of 2 cubes cannot be a double cube].

If p and r are odd, then we must have q=2 and consequently

p*r^3=8+s^3=(2+s)*(4-2s+s^2),

where the last 2 factors have GCD equal to 1 or 3.

If GCD=3 is the case, we get r=3, s=7.

If GCD=1 we have

1) 2+s=p, 4-2s+s^2=3+(s-1)^2=r^3

or

2) 2+s=r^3, 4-2s+s^2=p.

The first case requires a square and a larger cube at a distance of 3,
which seems unlikely to happen [See: http://www.research.att.com/~njas/sequences/A087285 .That is all I know. Therefore we cannot expect to have a^2+3=b^3].

Second case seems to have an infinity of solutions:

Take prime r.
See if s=r^3-2 is prime.
See if p=4-2s+s^2 is prime.

r = 439 s = 84604517 p = 7157924127594259
r = 619 s = 237176657 p = 56252766151342339
r = 2131 s = 9677214089 p = 93648472504985671747
r = 4801 s = 110661134399 p = 12245886666252218822407
r = 7237 s = 379031861051 p = 143665151691026507102503
r = 7591 s = 437418326069 p = 191334791980131168340627
r = 16267 s = 4304496906161 p = 18528693615141011845945603
r = 17839 s = 5676903560717 p = 32227234037669999498432659
r = 18679 s = 6517197260837 p = 42473860136648261419418899
r = 21997 s = 10643644593971 p = 113287170242746806160360903
r = 36319 s = 47907294649757 p = 2295108880658539862710859539
r = 38281 s = 56098315742039 p = 3147021029093388482560393447
r = 56167 s = 177191826009461 p = 31396943204566745348009491603
r = 59281 s = 208327481285039 p = 43400339458567858139802661447
r = 82039 s = 552155082225317 p = 304875234827245472962591299859
r = 82351 s = 558478722689549 p = 311898483696949056070788444307
r = 92461 s = 790452465768179 p = 624815100638992612907637439687
r = 92809 s = 799411294231127 p = 639058417344283905527507227879

Conclusion: I suspect that there are infinitely many solutions to the
problem: The two solutions mentioned in the problem formulation and the
above (most likely infinite) family of solutions.

***

Farideh wrote:

Q1: I think there exist many solutions for the equation p = (q/r)^3 + (s/t)^3.
Some of them are listed below.

1. p = 7157924127594259 q = 2 s = 84604517 r = t = 439
2. p = 56252766151342339 q = 2 s = 237176657 r = t = 619
3. p = 93648472504985671747 q = 2 s = 9677214089 r = t = 2131
4. p = 12245886666252218822407 q = 2 s = 110661134399 r = t = 4801
5. p = 143665151691026507102503 q = 2 s = 379031861051 r = t = 7237
...

In fact for the above class of solutions q=2, r=t, s = t^3-2 & p=s^2-2s+4.
We can easily find many such solutions.
By using this fact that the Diophantine equation x^3=y^2+3 has no integer solution,
we can easily prove that except for p=13(q=2,s=7,r=t=3) & p=19(q=5, s=3, r=t=2)
all other solutions are of the mentioned form.

Q2: We can easily prove that 13 is the smallest such p.

Q3: There is no a prime p such that q, r, s & t are four distinct primes.

Proof:
If p = (q/r)^3 + (s/t)^3 then p*r^3*t^3=q*t^3+s*r^3 so r|q*t^3 & t|s*r^3 we also
know that q,r,s & t are primes and q/r is different from s/t hence we
deduce that r = t and q<>s.

Some more explanations:

Since q/r & s/r=s/t are two distinct numbers we can deduce that q & s are two
distinct primes and we can suppose that q<s.

p=(q/t)^3+(s/t)^3 so p*t^3=q^3+s^3 and we there are following two
cases.

Case 1. If r=t=2, 8*p=q^3+s^3 hence q & s must be two distinct
odd primes and 8p=(s+q)(s^2+q^2-sq).
Since s^2+q^2-sq is odd we deduce that s^2+q^2-sq=p & s+q=8 so
the only solution for the case t=2 is p=19(t=r=2,q=3,s=5).

Case 2. If t(=r) is an odd prime, since p>2 and q^3+s^3 is odd
q must be 2. Hence we have (2+s)(4+s^2-2s)=p*t^3 since s,t & p
are distinct primes we have the following six cases.

I. 2+s=t^2 & 4+s^2-2s=p*t
II. 2+s=p & 4+s^2-2s=t^3
III. 2+s=t & 4+s^2-2s=p*t^2
IV. 2+s=p*t^2 & 4+s^2-2s=t
V. 2+s=p*t & 4+s^2-2s=t^2
VI. 2+s=t^3 & 4+s^2-2s=p

We can easily prove that:

1. For case I, p=13(t=r=3, q=2, s=7) is the only solution.

2. Since the Diophantine equation x^3=y^2+3 hasn't any integer solution, for the
case II there isa no solution.

3. For the cases III, IV & V there are no solutions.

So all other solutions are in the case VI and 13 is the smallest p.

***