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Still not right, Candledave!

Two of the integers sum to p(p+1)/2, etc.

But yes, the three integers sum to s.

Ha - you’re right. I shouldn’t try and help past 10pm!

Hi Candlewave and all other posters,
My brain (what there is of it!) is in a tizzy despite all the generous nudges and I am far off any solution to the equations. I have concentrated on clue VII but there seem to be too many combinations; I made the assumption that 2[k]60 (having subtracted 3 from the original figure) is a product of 16 and another four-digit number ending in 60 but to no avail; and further that two of the square terms on the LHS must probably be ninety somethings.
It's probably asking too much, but can a cryptic pointer to the thematic triples be given? I thought about Pythagorean triples, Fibonacci sequences and several other possibilities - but to no avail.
A fascinating puzzle but baffling me.

Unstuck - you are correct that two of the squares must be ninety something.

It's your assumption that s must end in 60 that's wrong (although I can see why you thought that).

S must end in 5 or 0 to get 16s ending in 0. Based on the size of the right hand term, and the fact that s must be triangular, there are only a fairly limited number of triangular numbers that s can be.

Many thanks SH for your swift reply confirming two of the squares must be in the nineties.
I'll give the puzzle another go tomorrow.

Thanks for a brilliant puzzle, Oyler.

I thought the preamble was perfectly fair - no need for any further explanation. Of course it's difficult to understand, but that is due to the convoluted nature of the puzzle, not because the explanation is poor.