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Pythagorean Quadruples
By Andras Cserep (P2357) on Sunday, April 9, 2000 - 09:45 pm:

I am doing some coursework at school on Pythagoras. I'm in year 10 and I understand his theory well. Our coursework is on Pythagorean triples which I have solved. As an extension, to get better marks, we were told to also find formulas for pythagorean quads. I have done that so my teacher told me to look at other 3D shapes which I could use. ( basically to calcuate the longest diagonal inside a 3D shape using pythagoras and using only whole numbers ) I can't find any shapes which I could use that could not be simplified into the pythagorean triples. If you understand what I've just said then please help if not I'll explain it again. Thanks.

By The Editor:

If you are reading this thread because you have been set the same coursework, please remember that any help you get from any source must be referred to in your write-up. This will not necessarily affect your mark, particularly where we have given only a hint.

In this case, the first reply gives just a hint, but the second reply gives a formula. If you want to try and find the formula for yourself, don't read the second one! If you do read it, you can say "I did some research on the internet and found this formula", and then gain credit by proving that it will always give a set of Pythagorean quadruples.

By Kerwin Hui (P1312) on Monday, April 10, 2000 - 02:46 pm:

Try the quadruples

(8,9,12,17)
(15,36,52,65)
...

Fun hunting,

Kerwin Hui

By Jeremy Burridge (T3006) on Tuesday, July 18, 2000 - 09:29 am:

Pythagorean 'quadruples', which are the 3D version of the triples, can all be generated by the following formula, where a, b, c and d are whole numbers.
If p = a2 + b2 - c2 - d2
q = 2(ad - bc)
r = 2(ac + bd)
and s = a2 + b2 + c2 + d2

then p2 + q2 + r2 = s2.

e.g. a=4, b=1, c=1, d=1 gives 15,6,10,19