Problems & Puzzles: Puzzles

Puzzle 498. p*s+q*r is never a square. Sebastian Martin Ruiz sent the following puzzle: Let p<q<r<s to be four consecutive prime numbers. p*s+q*r has a square free prime factor. Then p*s+q*r is never a square. Q. Is it true?

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Enoch wrote:

Squares type p.q+r.s are possible. Here is the first one; there must be others over the rainbow:

203233*203249=41306904017
203279*203293=41325197747
41306904017+41325197747=82632101764 with sqrt=287458

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Farideh wrote:

I'm sure that there is no proof for the assertion : All numbers of
the form p*s + q*r has at least one prime factor of power 1.

Because,

1. I think there is no main difference between numbers of the form p*s + q*r and
numbers of the forms p*q + r*s or p*r + q*s.

2. For each of the cases p*q + r*s and p*r + q*s there exist at least one number
which hasn't a prime factor of power 1.

Case p*q + r*s ( I wrote before) : {p, q, r, s} = {203233, 203249, 203279, 203293} ;
p*q + r*s = 82632101764 =2^2 * 143729^2


Case p*r + q*s : {p, q, r, s} = {8761, 8779, 8783, 8803};
p*r + q*s = 154229400 = 2^3 * 3^3 * 5^2 * 13^4.

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Carlos Rivera wrote:

a) Best numerical approximations to the asked question are:

p	q	r	s	p.s+q.r	m	m^2	Delta = p.s+q.r - m^2 < 10
5419	5431	5437	5441	59013126	7682	59013124	2
3	5	7	11	68	8	64	4
7	11	13	17	262	16	256	6
124309	124337	124339	124343	30916892230	175832	30916892224	6
421205627	421205633	421205641	421205677	354828389917320232	595674735	354828389917320225	7
461151253	461151323	461151347	461151371	425321086331599944	652166456	425321086331599936	8

b) Given p then q, r & s are away from p only in several possible distances given by the primes "rules" (unknown):

p
q=p+a
r=p+b
s=p+d

So p.s +q.r= p(p+d) + (p+a).(p+b) = (p^2+p.d) + (p^2+p(a+b)+ab) =2p^2+p(a+b+d)+ab

So, the Sebastian question is:

Can 2p^2+p(a+b+d)+ab = Ap^2 +B.p + C = m^2 be satisfied, if the coefficients A=2, B=(a+b+d) & C=a.b all even numbers restricted to the prime possible distances & p is prime?

Does this way of posing the question makes more light or shadow?

c) If p, q, r, s need not to be all four consecutive, but just three of them (in red below), here are some small solutions:

p	q	r	s	p.s+q.r	m
2	3	5	17	49	7
7	17	19	23	484	22
11	17	19	23	576	24

Accordingly, there must be some solutions for the Sebastian's question out there somewhere...

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