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Hi, rrobbo! The value for PRXX in the ratio in 4d is indeed 636. If you don't want to try all possible values of J in the other terms of the ratio (AJ+P+Q - I'm assuming that you have already solved for A, P and Q), then you could use Euclid's formula.

You can assume that 636 is equal to 2rs. It can't be r^2 - s^2, or that would make the other term equal to 2rs. Hence, both terms would be divisible by two, which contradicts the statement that the ratio is in its lowest terms, This means that rs is equal to 318.

You can factorise rs into two numbers in four different ways: 318*1, 159*2, 106*3, 53*6. However, for the first three factorisations, you will find that r^2 -s^2 has at least 5 digits, and 4d's answer has to have only four. So r is 53 and s is 6 in Euclid's formula, and you can work out what AJ+P+Q is by using the value you've calculated for r^2 - s^2, and hence get the value for J.

I suspect that no table of Pythagorean triples on the internet contains all the answers (those for 14a and 8d in particular), but the triple for 4d is on the table at http://www.tsm-resources.com/alists/PythagTriples.txt, which gives values for triples (a, b, c) where c goes as far as 10,000.

Many thanks both. Those comments explain a lot. I was avoiding using Euclid - after two weeks of holiday my brain is not quite ready for that yet and my maths is wavering these days anyway. It will probably do me no harm to revisit such topics I covered in my A levels all those years ago....but only after I have engaged my true analytic brain again tomorrow for work. This holiday one is rather loathe to stop relaxing 8-)

Oh man.......I had looked at that SS but not in column B!!!!

This list also misses a couple of triples, including 4d, but I did find it very useful nethertheless:

http://web.gccaz.edu/~wkehowsk/187-Precalculus-08-09-Su/triples.pdf

I really must be missing something here. I have Z = 2 and P = 3. With 1ac, I appreciate that H = r + s, B – H = 2rs, and that r = s + 1. Consequently, I have H = {5, 7, 11, …, 43}, since H = 47 yields a value for c that is too large for the space available. Corresponding values of (B – H) are {12, 24, 60, …, 924}. In total, I seem to have, then, 12 possible pairs that can combine in a Pythagorean triple, and I simply cannot see how to eliminate any of them. I’d be very grateful for any help with this, please.

...and almost as soon as I post, I see that I have to combine the info in 1ac with that in 4ac which, I think, means that H is 11 or 13. Perhaps if I then consider 10ac, further pruning may be possible. I'd still be grateful for an indication that I'm now on the right lines. Thanks.

h is indeed 11 or 13

Hi, carpox1! Yes, H is 11 or 13. I got to this stage by using Excel to calculate all the possible combinations of H and B in 1a, and found only four combinations that gave a Pythagorean triple. Then I did the same in 4a for all possible combinations of the four values of H given from 1a and the remaining prime numbers that could be Y. This reduced the possibilities to two sets of values for H, B and Y, with H being either 11 or 13. Unfortunately, checking the values of H and Y in 31a did not eliminate one of the sets, so I still had two sets of values for H, B and Y. Brute computation of all possibilities in 10a gave me a value of T corresponding to each set of values of H, B and Y, so I now had two possible sets for H, B, Y and T. Because I already knew Q and V, I was able to use 11a to eliminate one of the two sets, and also calculate D, which gave me 11 unambiguous values (A, B, D, H, P, Q, R, T, V, X, Y) after 9 clues.

There are probably more elegant solution paths using Euclid's formula, but I find brute computation using Excel is often quicker in this sort of puzzle.

Many thanks to tatters and Wintonian: very helpful and reassuring.

Pretty sure Z isn't 2. 9d confirms that for me I think, but being that I am now stuck half way through with a value for E that simply does not play ball, I could well be wrong! Still holding Euclid at bay but I think the time is nearing....