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We have this theorem (https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem) which gives conditions on positive integer $n$ such that $n=a^2+b^2$. I was wondering if there exists any such $n>1$ which is also a perfect cube? In other words, we seek the solutions of Diophantine equation $a^2+b^2=z^3$.

We see $a=b=z=2$ is one of the solutions. I am interested in other positive non-trivial solutions with $a\ne b$.

John Omielan
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ersh
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    Any $n$ whose prime factorization has no prime $3$ mod $4$ to an odd power is a sum of two squares. This can be applied when $n=z^3$ to get restriction on $z^3,$ then on $z.$ – coffeemath Jan 01 '19 at 02:08
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    https://math.stackexchange.com/questions/334839/variation-of-pythagorean-triplets-x2y2-z3/724370#724370 – individ Jan 01 '19 at 06:57

3 Answers3

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Consider $b = ka$ for some integer $k > 1$. Then we have

$$a^2 + b^2 = a^2 + \left(ka\right)^2 = \left(1 + k^2\right)a^2 = z^3 \tag{1}\label{eq1} $$

Now, if $z = a$, \eqref{eq1} would be true if

$$k^2 + 1 = a \tag{2}\label{eq2} $$

Thus, if $k = 2$ for example, then $a = 5$ and $b = 10$ giving

$$5^2 + 10^2 = 5^3 \tag{3}\label{eq3} $$

There are, of course, many other such similar examples. If you wish for $a \neq z$ as well, then you could also have, for example, that $z = ga$, for an integer $g \gt 1$, so $1 + k^2 = \left(g^3\right)a$, but there are no solutions for certain cases, such as $g = 2$.

More generally, if you wish to have other restrictions, such as that $\gcd\left(a, b, z\right) = 1$, then consider what coffeemath wrote in a comment to the question. In particular, any number $n$ is a sum of two squares if and only if all prime factors of $n$ which are $\; 3 \mod 4 \; $ have an even exponent in the prime factorization of $n$. This is stated and proven in Which Numbers are the Sum of Two Squares?. Thus, any $z$ with all prime factors which are $\; 3 \mod 4 \; $ having an even exponent in its prime factorization will work. My example of $5$ is basically the simplest such case involving positive integers.

As Mike Miller pointed out in the comments to this answer, there is a formula for the number of representations of a number as the sums of squares at Sums of squares function, although you might need to remove cases where it provides $a$ or $b$ to be $0$ as the question specifically is looking for only positive integer solutions.

John Omielan
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    This is absolutely what I was looking for! Thanks! – ersh Jan 01 '19 at 02:20
  • @ersh You are welcome. I have expanded on what coffeemath wrote to give a more general solution. – John Omielan Jan 01 '19 at 02:40
  • If I fix one of the variables, say $z$ (or any other variable), can I get all the parametric solutions or at least some family? For example, what would be the family of solutions to your special example $a^2+b^2=5^3$? – ersh Jan 01 '19 at 03:02
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    There is also a nice formula here for the number of representations of $z$ as $a^2 + b^2 = z$, so you can even count solutions. Getting a formula to actually list out the solutions is much less likely. –  Jan 01 '19 at 03:03
  • @MikeMiller Thanks for the comment and link. I have added your information to my answer. – John Omielan Jan 01 '19 at 03:11
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    John-- How is your "more general" solution more general than that of my comment? In both cases the primes $3$ mod $4$ should appear to even power. – coffeemath Jan 01 '19 at 03:12
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    @coffeemath Sorry, I misread your comment. I have corrected my answer accordingly. – John Omielan Jan 01 '19 at 03:16
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Above equation shown below:

$a^2+b^2=z^3$ -----(A)

I came across the parametric solution of equation (A) on the internet

and have written it below:

$a=s(s^2-3t^2)$

$b=t(3s^2-t^2)$

$z=(s^2+t^2)$

For $(s,t)=(3,2)$, we get:

$9^2+46^2=13^3$

Sam
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    Sam-- Could you link to internet article where parametrization is given? – coffeemath Jan 01 '19 at 06:30
  • @coffeemath, I found this (https://www.emis.de/journals/GM/vol13nr2/andrica/andrica.pdf) where reference is also given to Sam's solution and and what S.I posted. – ersh Jan 01 '19 at 16:30
  • Sam-- That solution of your link says it's an infinite parametrized solution, not claiming it is all of them. – coffeemath Jan 01 '19 at 17:48
  • @cofeemath. The solution (identity) could be a general solution, but I have not come across a proof yet. Untill the day somebody comes up with a numerical solution that is not satisfied by the identity we may have to assume it is so. Just as, [(m^2-n^2)^2+(2mn)^2= (m^2+n^2)^2] is a general solution to the phytogorous equation – Sam Jan 01 '19 at 23:17
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$$ a x^2 + b y^2 = z^3 $$ $$ a x^2 + b y^2 = z^5 $$ $$ a x^2 + b y^2 = z^7 $$ $$ ... $$ https://youtu.be/WycaoCnKcU0

S. I.
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