We know that $5^2 = 3^2 + 4^2$. But it is possible to show $a^2$ as the sum of three or more distinct squares?
Something like that exist? $a^2 = b^2 + c^2 + d^2$, where $b$, $c$, $d$ are distinct natural numbers.
Thanks a lot!
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$\begingroup$ $a^2=1^2+1^2+\cdots+1^2$ is one way. If $a\geq 2$, you can swap four of the $1$'s for a $2$. Likewise for $3$, $4$, or any other number. $\endgroup$– ArthurOct 27, 2014 at 14:30
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$\begingroup$ It is possible to solve any such equation. Even set some conditions. Only question is, what exactly should be sought. math.stackexchange.com/questions/514075/… $\endgroup$– individOct 27, 2014 at 15:03
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3 Answers
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Legendre's three-square theorem states that every number that is not of the form
$$4^n(8m+7)$$
can be written as a sum of exactly three squares. Since most (all?) squares are not of this form, the answer to your question is Yes.
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2$\begingroup$ Yep, all of them. Easy to show that for negative values of n this term can't be an integer. For positive values of n the term $4^n$ is a square number, hence $8m+7$ has to be a square number which is not possible. $\endgroup$– Galc127Oct 27, 2014 at 14:53
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The complete solution for your particular case ($n=3$) was given by Catalan — just about any elementary number theory text will give the general integer parameterization of the primitive solutions, which [assuming $b$ is odd] is \begin{align} a &= p^2+q^2+r^2+s^2, \\ b &= p^2+q^2-r^2-s^2, \\ c &= 2(ps+qr), \\ d &= 2(pr-qs). \end{align}
The complete answer for any $n \ge 2$ is given in several papers. A good introduction is Bradley’s paper Equal Sums of Squares. For a more technical matrix-based approach, read Barnett and Mendel’s paper On Equal Sums of Squares.
answered Oct 27, 2014 at 15:48
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1$\begingroup$ Those references are excellent. How did you know about those? $\endgroup$ Aug 24, 2019 at 1:10
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1$\begingroup$ @AmateurMathPirate: ESOS equations have been a focus of mine for over a decade. $\endgroup$ Aug 24, 2019 at 16:56
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Yes, they do exist! An example is $(2, 3, 6, 7)$. $\sqrt{2^2+3^2+6^2}=7$. These are called Pythagorean quadruples. They exist in the form $(a,b,c,d)$. There is a Wikipedia artible about this: click here.
A Pythagorean quadruple is mostly used to calculate the space diagonal ($d$) of a cuboid, which is equal to $\sqrt{a^2+b^2+c^2}$, where $|a|$, $|b|$ and $|c|$ are the length, width and height of the cuboid. Here is an example:
(source: smpn1bontang.org)
$QW$ is the space diagonal, which is equal to $\sqrt{PS^2+RS^2+WS^2}$
