Find all primes p and q such that p + q =
(p − q)3.
Answers
Answer:
The following has been unanswered in art of problem solving and other forums for months.
Find all primes p,q such that
p3+p=q7+q
One solution is (5,2) and it has been computer checked that there is no other solution till 107.
I am really curious to see a complete solution.
number-theory prime-numbers
Where does this particular equation come from? We could ask whether a particular Diophantine equation has a solution in primes (when finding an integral solution is hard enough), but why is that interesting? – anomaly May 7 '17 at 22:40
Well , I don't know if there is anything interesting to be found from it but I wanted to see if it can be solved using highschool olympiad mathematics.As i said many have tried but all failed so I posted it here as a last resort. – Andreas Ch. May 7 '17 at 22:49
If it's an unsolved problem in maths, I think you could be better off on MathOverflow, though I'd hardly hope for a solution on a whiff on any forum. – AspiringMathematician May 7 '17 at 22:51
you will have a prime pair solution when 1+p2+q6=0modpq – Ahmad May 7 '17 at 23:58
answer1
11
+100
Assume we have another solution, so that q≥3. We know that p(p2+1)=q(q2+1)(q4−q2+1). We also have p>q2+1. In fact, otherwise p≤q2+1 and then
q7+q=p3+p≤(q2+1)3+q2+1=q6+3q4+4q2+2,
and so q7<q6+3q4+4q2+2<2q6, which is impossible if q≥3.
Since we also have p>q, it follows that p|(q4−q2+1) and so there exist a natural number k such that pk=q4−q2+1. But then
p(p2+1)=q(q2+1)pk,
and so p2+1=q(q2+1)k. This implies
p2k2+k2=q(q2+1)k3.
Since p2k2=(q4−q2+1)2, we arrive finally at
(q4−q2+1)2+k2=q(q2+1)k3,
which we rewrite as
(q3+q)(q5−3q3+6q)−8q2+1+k2=(q3+q)k3.
So we obtain
(1)(q3+q)(k3−q5+3q3−6q)=k2+1−8q2.
It follows that k2+1≡0modq.
Let y∈{0,1,…,q−1} be such that k≡ymodq.
If y<q−1−−−−√ or q−y<q−1−−−−√, then 0≤y2<q−1 or q(2y−q)≤y2<q(2y−q)+q−1, so in particular it is impossible that k2≡y2≡−1modq.
Now, since k3≡−k≡−ymodq, for any N∈N we have |k3+qN|≥q−1−−−−√. So, from (1), we deduce that
|1+k2−8q2|=(q3+q)|k3+q(3q2−q4−6)|≥(q3+q)q−1−−−−√.
By hand one verifies that q=3, q=5 and q=7 are not solutions, so we can assume q>8, hence q3>8q2 and so necessarily k2+1>8q2 and then |1+k2−8q2|=k2+1−8q2<k2.
So we arrive at
k2>q3q−1−−−−√.
But then k12>q18(q−1)3 and so
(p2+1)12=k12q12(q2+1)12>q18(q−1)3q12(q2+1)12>q54(q−1)3.
On the other hand p>10 implies 2p8>(p2+1)4, and so
(p2+1)12=(p2+1)8(p2+1)4<2(p2+1)8p8=2(p3+p)8=2(q7+q)8.
Putting the inequalities together, we obtain finally
2(q7+q)8>q54(q−1)3,(∗)
which is impossible, since the left hand side is of order 56 and the right hand side has order 57.
Detailed proof that (*) is impossible:
Since q>10, we have that q−1>910q and since (910)3>12, we obtain (q−1)3>q32. Hence (*) implies
4(q6+1)8>q49,(∗∗)
On the other hand, for any α>106 we have 2α8>(α+1)8, for example, using that (1+1α)8<(1+10−6)8<2.
In particular, for α=q6>106, we have 2(q6)8>(q6+1)8, and so from (**) we obtain
8q48=4(2(q6)8)>4(q6+1)8>q49,
which is impossible, since q>8.
Note: The proof uses crucially that p is prime when deducing that p divides q4−q2+1. However, the condition that q is prime is not needed.
Well, (1+1α)8<(1+1255)8<2 is simpler... – guest May 12 '17 at 2:36
Such a simple solution! Me and the creator of this problem congratulate you and thank you for your help. – Andreas Ch. May 12 '17 at 10:17
Could you highlight the specific point(s) in this proof where the assumption that p and (especially) q are primes is used? – Barry Cipra May 12 '17 at 13:47
I only used that p is prime, when I deduced that p divides q4−q2+1, and used that q>8, checking primes 3,5,7 by hand, for which it was not really necessary for q to be prime. – san May 12 '17 at 13:59
My edit was only for a small typo. Following (1), in the sentence "So,from (1), we deduce that....", the term −6q appeared as −6. – DanielWainfleet May 12 '17 at 15:45
Step-by-step explanation: