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], 2529, 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 almostsolutions
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.
***
