My reputation is at this moment at $1600$.

I did some experimenting with $1600$ and obtained the following:

Evidently, it is a perfect square $1600=40^2$

Also, it is a hypothenuse of a Pythagorean integer-triple triangle $1600=40^2=32^2+24^2$.

Also, it can be written as the sum of four non-zero squares $1600=20^2+20^2+20^2+20^2$

So, $1600$ is a perfect square, a hypothenuse of a Pythagorean integer-triple (so can be written as a sum of two non-zero squares), and a sum of four non-zero squares.

Is there an infinite number of numbers like $1600$? Can you find some more?

Edit 1: It is also a sum of $4$ non-zero positive cubes: $1600=8^3+8^3+8^3+4^3$

Edit 2: It is also a sum of powers from $1$ to $4$, as we see $1600=7^1+9^2+6^3+6^4$

  • $\begingroup$ You mean are there an infinite number of numbers satisfying those three things? $\endgroup$ – Dave Dec 30 '17 at 5:29
  • $\begingroup$ @Dave Yes, right that. $\endgroup$ – user480281 Dec 30 '17 at 5:29
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    $\begingroup$ See if this helps you: eulerarchive.maa.org/docs/translations/E228en.pdf $\endgroup$ – Rohan Dec 30 '17 at 5:31
  • $\begingroup$ Your edit has changed the question completely. The previous version was homogeneous, so we could just scale one solution to find more. The edit makes that fail. -1 $\endgroup$ – Ross Millikan Dec 30 '17 at 5:39
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    $\begingroup$ Your second edit is very trivial, though, since $n=(n-k+1)^1+1^2+\cdots+1^k$. $\endgroup$ – Clayton Dec 30 '17 at 6:04

Yes, there are an infinite number of them. Given your result for $1600$ we can say that $1600k^2$ is another one. It is a square, it is $(32k)^2+(24k)^2$ and it is $4\cdot (20k)^2$. We can use the parameterization of Pythagorean triples to get others. If you choose $m,n$ relatively prime and of opposite parity, they generate a primitive Pythagorean triple $m^2-n^2, 2mn, m^2+n^2$, so if we choose $m,n$ as legs of a Pythagorean triangle the hypotenuse of that triangle will be a square. As an example, let $m=4,n=3$, which gives the triangle $7,24,25$. The number $25$ is a square, the sum of two squares, and the sum of four squares as $16+4+4+1$. Again you can multiply it by any square.

With the edit, we can still say that $1600k^6$ is a solution by the same reasoning as above plus $1600k^6=3\cdot(8k^2)^3+(4k^2)^3$.


Yes. We know that $5^2=3^2+4^2$ and, if we consider $k\in\mathbb{N}$ and a semejant triangle for this with scale $2k$ then $(6k)^2+(8k)^2=(10k)^2$.

The number $(10k)^2$ is our candidate. In fact, its is a square number, its a hypothenuse of a pythagorean integer-triple.

Also, $$(10k)^2=(2\cdot 5k)^2=4\cdot(5k)^2=(5k)^2+(5k)^2+(5k)^2+(5k)^2$$

is the sum of four square numbers.


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