Problems & Puzzles: Problems

Problem 70. Set of integers whose sum of any two is a perfect square.

Here we ask for K distinct positive integers such that the sum of any two of them is a perfect square.

The record is K=5 and was obtained in 1971 by Dr. J. L. Nicolas(*).

{7442, 28658, 148583, 177458, 763442}

7442 + 28658 = 190^2
7442 + 148583 = 395^2
7442 + 177458 = 430^2
7442 + 763442 = 878^2
28658 + 148583 = 421^2
28658 + 177458 = 454^2
28658 + 763442 = 890^2
148583 + 177458 = 571^2
148583 + 763442 = 955^2
177458 + 763442 = 970^2

Perhaps 46 years is time elapsed enough to give another try to this challenge.

Q1. Find a set of K=6 distinct positive integers such that the sum of any two is a perfect square.

Q2. Find a set of K distinct prime integers such that the sum of any two is a perfect square, for K>2 or prove that this is impossible.

(*) See A.R. Thatcher, “Five Integers Which Sum in Pairs to Squares,” Mathematical Gazette 62:419 [March 1978], 25-29, or http://www.jstor.org/stable/3617620, or this local pdf document.

Contributions came from Dmitry Kamenetsky & Jan van Deleden

***

Dmitry wrote (9 Set, 2017):

Ajai Choudhry has found six positive integers, but two of them repeat:

3694388882, 3694388882, 60445225682, 42248104082, 102804712082, 254645020559.

He has also found six positive integers, but one of them is negative

339323777731946898, 1393697157060854002, 2146648434867118098, 8397374854916636127, 12982930841197954098, −303704776155745998.

Attached is his paper and it should be a good starting point for work on this problem.

...

(15 Set, 2017)

1. I have found almost-solutions with 6 numbers. In the following only one of the sums is not a perfect square:
9122 104447 208034 348482 1295042 2307362
36488 417788 832136 1393928 5180168 9229448
82098 940023 1872306 3136338 11655378 20766258
145952 1671152 3328544 5575712 20720672 36917792
334226 515858 641918 3155198 3835538 7629458

2. I have not found any prime solutions for primes under 20 million.

***

Jan wrote (16 Set, 2017)

Q2: If there are two odd primes p,q with the sum equal to an even square, we must have that p+q=0 mod 4. So these primes must have a different residue mod 4, so p is unequal to q. Adding another odd prime (K>2) would have either p+r or q+r unequal to 0 mod 4. So the number of odd primes is limited to 2.

One could try to achieve K=3 by adding the prime 2 to a set of two odd primes.

Suppose p=1 mod 4 and q=3 mod 4:

2+p=a^2 (3 mod 4)

2+q=b^2 (1 mod 4)
But all odd squares must be 1 mod 4, so this is not possible. The other situation is covered in the same fashion.

There is only one option to achieve K>2:  The set of primes {2,2,2,…} which was excluded explicitly.

*** Records   |  Conjectures  |  Problems  |  Puzzles  Home | Melancholia | Problems & Puzzles | References | News | Personal Page | Puzzlers | Search | Bulletin Board | Chat | Random Link Copyright © 1999-2012 primepuzzles.net. All rights reserved.