Problems & Puzzles:
p = (q/r)3 + (s/t)3
Patrick Capelle sent the
We are interested by prime numbers p such that p = (q/r)
, where q,r,s,t are prime numbers and q/r is different from s/t.
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
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.
Can you give other solutions for p ?
What's the smallest p? Is it 13? Can you justify your answer?
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
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:
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
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
2) 2+s=r^3, 4-2s+s^2=p.
The first case requires a square and a larger cube at a distance of
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
problem: The two solutions mentioned in the problem formulation and
above (most likely infinite) family of solutions.
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 &
We can easily find many such solutions.
By using this fact that the Diophantine equation x^3=y^2+3 has no
we can easily prove that except for p=13(q=2,s=7,r=t=3) & p=19(q=5,
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
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
know that q,r,s & t are primes and q/r is different from s/t hence
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
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.