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Hi Just filling in. I know this is a crossword Forum but Ive been submitting since before chrise left school so can I pinch a few bytes with a problemo that Ive been wrestling with for months cant I ?

P, Q and R are integers and
P + Q , P + R, Q + R, P - Q, P -R and Q - R are all perfect squares. Probably a job for QBasic or something else. Any ideas or comments very welcome Thea(o)

I have the answer but maybe it is better to just give you a couple of humungous clues.

Q and R are both single digit non-zero integers and Q+R=0

P

Oops

P is less than 20

Let’s assume that P, Q and R are all different integers. If not, you can have trivial solutions such as P=Q=R=0, P=Q=R=2, and so on (zero is both an integer and a perfect square).

Tatters has identified what I think is the solution with different integers that represents the values of P, Q and R closest to zero. However, there is an infinite number of solutions. Any Pythagorean triplet {a, b, c}, where a, b and c are all positive integers, and a^2+b^2=c^2, generates a solution where b is even. The solution is

Q=(b^2)/2
R=-(b^2)/2
P=a^2+Q=c^2-Q

If a is even, the solution is:

Q=(a^2)/2
R=-(a^2)/2
P=b^2+Q=c^2-Q

In any Pythagorean triplet, at least one of a and b must be even. If both are even, a triplet generates two solutions. For example, the triplet{6, 8, 10} leads to the following solutions:

P=82, Q=18, R=-18
P=68, Q=32, R=-32.

Are there any solutions where all three of P, Q and R are positive and distinct?

Hi wintonan and tatters - Thanks very much for your posts - I shall resume the ANALYSIS !!!! . As mentioned I have recently downloaded Qbasic so might get too many solutions as the industrious chemist said. Will try to remember you both with posts if my programming is up tp scratch. Cheers Have a nice weekend ( See you at the Jumbo)) THEA(O)