Problems & Puzzles:
Puzzles
Puzzle 109. a_{1}<a_{2}<...<a_{k} such
that...
Here
you are asked to find k distinct integers
a_{1}<a_{2}<...<a_{k}
such that the sum of any two of them is a square.
The
problem becomes a rather interesting one when you are asked to find the
particular set of k values if a_{k}
is conditioned to be the smallest prime possible number, for each
k=>2
Here
are the solutions that I have found for the first k values:
k 
Set
of a_{k}
values 
2 
{1,
2} 
3 
{4,
20, 29} wrong!!!...
{4,
4, 5} found by Imran Ghory,
30/9/2000 
4 
{622,
626, 818, 1583} 
5 

6 

7 

Questions:
1.
Can you extend the above table for k=
5, 6 & 7
2.
Can you demonstrate why for k=>3 all the numbers in the set must be
even numbers except the prime number
________
n. b.: This
puzzle is an extension of one published by Frank Rubin in one of
his extremely interesting puzzles for fun's pages.
But the less restricted problem (a_{k}
is neither prime nor minimal) "...goes back to T. Baker who
found (1839) five integers whose sum in pairs were squares and C. Gill
who found (1848) five whose sum in three were squares" . In 1972 Jean
Lagrange found one solution with 6 integers. Erdös & Leo Moser
ask (when?) if exist solutions for any quantity (k) of integers.(cfr. D15,
pp. 170171, R. K. Guy, UPiNT")
Chris Nash & Jud McCranie has
solved independently the question 2; their respective argument is the
same:
"Think in terms of 'mod 4'. All perfect
squares leave a remainder of zero or one on division by 4. Now suppose
two of the three numbers are odd. Their sum is even, and to be an even
square, it must be a multiple of 4. So one of the odd numbers leaves a
remainder of 1, the other a remainder of 3, after division by 4. So say
a=1 mod 4, b=3 mod 4. It's quick to check the possible values of c mod
4:
c=0 mod 4: b+c=3 mod 4, cannot be square
c=1 mod 4: a+c=2 mod 4, cannot be square
c=2 mod 4: a+c=3 mod 4, cannot be square
c=3 mod 4: b+c=2 mod 4, cannot be square
So if a and b are both odd, there is no solution for c. So we're
forced to conclude at most one of the numbers can be odd."
***
Giovanni Resta wrote (Nov. 2004)
I found this set for K=5.
25822, 31298, 43778, 114626, 351863
The sums are
25822+31298=74^2
25822+43778=134^2
25822+114626=298^2
25822+351863=571^2
31298+43778=274^2
31298+114626=382^2
31298+351863=619^2
43778+114626=398^2
43778+351863=629^2
114626+351863=683^2
***
