It is, of course, quite possible for two consecutive numbers both to be the sum of two squares; for example 17=4^2+1^2, and 18=3^2+3^2.

Now although x is not itself a sum of 2 positive squares, each of x-3,x-2,x-1,x+1,x+2 and x+5 is a sum of two positive squares, say a^2, b^2, c^2...etc. such that 8 of the numbers a,b,c...etc. form consecutive members of an arithmetic sequence.

What is x?

All perfect squares are congruent to 0 or 1 (mod 4), so the sum of two squares can never be congruent to 3 (mod 4). Since x – 3, x – 2 and x – 1 are sums of two squares while x is not, it follows that x = 3 (mod 4), so that a computer search need only try x = 7, 11, 15, ....

If my maple code is correct, the first value of x for which x – 3, x – 2, x – 1,
x + 1, x + 2, x + 5 are all sums of two squares is x = 291, which happens to provide 8 consecutive terms of an arithmetic sequence, as shown below.

x – 3 = 288       = 12^2 + 12^2
x – 2 = 289       = 8^2 + 15^2
x – 1 = 290       = 11^2 + 13^2
x + 1 = 292       = 6^2 + 16^2
x + 2 = 293       = 2^2 + 17^2
x + 5 = 296       = 10^2 + 14^2

Sequence:         10, 11, 12, 13, 14, 15, 16, 17.

Posted by Harry on 2011-10-18 19:19:30