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Can anyone think of 5 positive integers, a,b,c,d and e such that

a^2 + b^2 + c^2 + d^2 = e^2 ?????

[ There is more than one correct answer ] !!!!!!

(Also....and I should have been more specific....the integers are all different ) !!!!

CPhill Oct 8, 2014

#5**0 **

Ding!! Ding!! Ding!!!......we have a winner!!

kitty<3 !!!!

I came up with........

1 , 2 , 8, 10, 13

Note, that any positive integer multiples of these answers "work," too. (Just like in the Pythagorean "Multiples")

[Mmmm...I wonder if there is a general "formula" for generating the "correct" values??? ]

CPhill Oct 10, 2014

#6**+5 **

Here's a way of generating more sets of values:

Choose a Pythagorean triple, say 3, 4, 5. Let a = 3m, b = 4m, c = 3n, d = 4n, and e = 5√(m^{2} + n^{2})

a^{2} + b^{2} + c^{2} + d^{2} = (3^{2} + 4^{2})m^{2} + (3^{2} + 4^{2})n^{2} = (3^{2} + 4^{2})(m^{2} + n^{2}) = 5^{2}(m^{2} + n^{2}) = e^{2}

Now choose m and n to be part of another Pythagorean triple, say m = 5 and n = 12, and we have e as an integer (=5*13).

So a = 3*5 = 15; b = 4*5 = 20; c = 3*12 = 36; d = 4*12 = 48; e = 5*13 = 65;

$${{\mathtt{15}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{20}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{36}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{48}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

$${{\mathtt{65}}}^{{\mathtt{2}}} = {\mathtt{4\,225}}$$

(I don't remember seeing Chris's last line when I gave my previous answer. Hmm!)

.

Alan Oct 10, 2014