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I was presented with the following problem:

Prove that there exist solutions to $x^2+y^2=z^n$ for all $n$, with $x,y,z, n \in \mathbb{N}$

I showed that by taking any Pythagorean triple $x^2+y^2=z^2$ and multiplying by $z^{2(n-1)}$ we get $(z^{n-1}x)^2+(z^{n-1}y)^2=(z^2)^n$, which allows me to generate solutions easily for any value of $n$. I managed to find several similar questions on this site, such as this one concerning the specific case $n=3$. I notice that all of these questions take a similar approach and start with a Pythagorean triple and use it to generate general solutions.

Is there a way to prove the statement (or better yet provide solutions to the equation) without first relying on Pythagoras?

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acernine
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    N is a natural number too; I just edited the question to reflect this. – acernine Dec 26 '16 at 11:45
  • Since $n \in \mathbb{N}$, infinitely many solutions may be generated when $n=1$. – projectilemotion Dec 26 '16 at 11:49
  • the issue is not correct. I must say that if any decisions do not have such a form. $z=p^2+s^2$ – individ Dec 26 '16 at 11:52
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    I suppose $x,y,z$ most not be zero, otherwise it's trivial –  Dec 26 '16 at 12:20
  • There are infinitely many primitive solutions (that is $\gcd(x,y,z)=1$). Take two relatively prime integers $p$ and $q$ such that $p-q$ is odd. Choose $x$ and $y$ such that $$x+yi=(p+qi)^n.$$ Then, $$x^2+y^2=z^n,$$ where $$z=p^2+q^2.$$ By the choices of $p$ and $q$, $\gcd(x,y,z)=1$. – Batominovski Oct 20 '19 at 10:55

2 Answers2

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The solution $(x,y,z)=(0,1,1)$ works for all $n$.

If you don't want to allow $0$, then let $x,y\in\Bbb{N}$ be such that $$x+yi=(1+2i)^n.$$ Then $$5^n=((1+2i)(1-2i))^n=(1+2i)^n(1-2i)^n=(x+yi)(x-yi)=x^2+y^2.$$

Alternatively, if $n$ is odd let $m:=\frac{n-1}{2}$ so that $$(2^m)^2+(2^m)^2=2^n.$$

Finally, less constructively, a theorem of Gauss tells us that if an integer is not divisible by any prime congruent to $3$ modulo $4$, then it is a sum of two squares. Hence solutions exist for any $n$.

Servaes
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    Question says x,y,z belong to N. 0 does not belong to N. – Swapnil Rustagi Dec 26 '16 at 12:26
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    I've rarely encountered an instance of $\Bbb{N}$ not containing $0$ outside this forum, but as it does make the question somewhat less trivial I'll updat my answer. – Servaes Dec 26 '16 at 12:29
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    The standard ISO 80000-2 (International Organisation for Standardization) includes $0 \in \mathbb{N}$. However, other definitions suggest that $1$ is the smallest element of $\mathbb{N}$ – projectilemotion Dec 26 '16 at 12:33
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    @Servaes. In this forum the consensus is that $\mathbb N$ is the positive integers, which is something I have reluctantly accepted. – DanielWainfleet Dec 27 '16 at 00:41
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    In that case, I suggest you use $\mathbb{Z}^+$ instead. It is significantly less ambiguous. – projectilemotion Dec 27 '16 at 01:39
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    For me, $0\in\mathbb{N}$. – user5402 Dec 29 '16 at 14:13
  • @inéquation I have friends who work at the Kahn Academy. They are math tutors, and they teach that $0\notin \mathbb{N}$ and that it is instead a whole number. I can't disagree, can I? If it is stated by the book... – Mr Pie Apr 09 '18 at 11:31
  • @DanielWainfleet yes, I have been taught that $\mathbb{N} = \mathbb{Z}^+$. – Mr Pie Apr 09 '18 at 11:33
  • @user477343 I don't understand why can't you disagree. $\mathbb{N}={0,1,2,\ldots}$, $\mathbb{Z}+=\mathbb{N}$ and $\mathbb{N}^*={1,2,3,\ldots}=\mathbb{Z}+^*$. – user5402 Apr 09 '18 at 17:57
  • @user477343. The meaning of the symbol $\Bbb N$ varies by country, by era, and by author. On this site the prevailing use is that $\Bbb N=\Bbb Z^+,$ which I have reluctantly acceded to. – DanielWainfleet Apr 09 '18 at 20:56
  • @inéquation $0$ is not a positive integer, because $0=-0$. Thus, $0\notin \mathbb{Z}^+$.... but thinking about it more, $0 \not< 0$ either so it really all depends, and in one way or another, I guess you cpould be right. – Mr Pie Apr 11 '18 at 05:40
  • @DanielWainfleet thank you for that note. I believe I had a very narrow-minded perspective on things. – Mr Pie Apr 11 '18 at 05:42
  • @user477343 I use the term positive in the bourbaki sense $\geq 0$. A natural number is a number that is used to count things: I have 1 cellphone, 2 bikes, 0 airplanes. In this sense, $0$ is just as natural as $1$ or $2$; it is the number of elements in a finite set (if there is no bijection between a set and any of its subset, we call it finite). – user5402 Apr 11 '18 at 11:51
  • @inéquation so would then the negative integers be $\leqslant 0$? I guess not if you separate the definition of the natural numbers from the positive integers, i.e. $\mathbb{N}\neq\mathbb{Z}^+$. I guess I could just say that $0\in \mathbb{N}$ and $0\notin\mathbb{N}^+$, but now $\mathbb{N}=\mathbb{W}$ which is the set of whole numbers... How I look at it is that when kids learn to count, they start from $1$ and not $0$. But sometimes they do. Like starting a week from Sunday rather than Monday. But for me personally, I treat the natural numbers as counting numbers starting from $1$ and not $0$. – Mr Pie Apr 12 '18 at 13:40
  • @inéquation briefly, there's no right or wrong definition as it's up to personal interpretation, I guess. Here I use $\mathbb{N}^+$ instead of $\mathbb{N}^*$. – Mr Pie Apr 12 '18 at 13:45
  • @user477343 Definitions are not right are wrong. Just be clear about which definition you are using whenever there might be confusion, as with $\Bbb{N}$. – Servaes Apr 12 '18 at 13:48
  • @Servaes ok. It's just that different books say different things. Like the use of my set notation was frowned upon in one of my questions I posted, but I was taught that it was completely fine. I will definitely be clear about things like this next time. Thanks :) – Mr Pie Apr 12 '18 at 13:52
  • @user477343 Yes negative means $\leq 0$. What do you mean by a whole number? The notation $\mathbb{N}$ is almost universal in many languages whereas $\mathbb{W}$ is far from it (it's the first time I see it). – user5402 Apr 12 '18 at 19:38
  • @inéquation I was taught that $\mathbb{N}={1,2,3,\ldots}$ and $\mathbb{W}=\mathbb{N}\cup{0}$. So $0\notin\mathbb{N}$ but $0\in\mathbb{W}$. Though I would definitely agree that $\mathbb{N}$ is more universal. – Mr Pie Apr 13 '18 at 02:20
  • @user477343 The concept of "whole number" is only limited to english. In french for example, "whole number" and "integer" mean the same thing: "nombre entier". I use $\mathbb{N}$ in the Bourbali sense ($0\in\mathbb{N}$). Other users on this site don't. – user5402 Apr 13 '18 at 12:14
  • @inéquation Apologies, I didn't realise. Do you think there should be a universal consensus on what definitions should be, such as for $\mathbb{N}$, in order to avoid confusion? – Mr Pie Apr 13 '18 at 12:44
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    @user477343 I think there should be a universal consensus on $\mathbb{N}$, on the decimal separator, on the open interval notation ($]a,b,[$ or $(a,b)$ like an ordered pair), positive ($\geq 0$ or $>0$), ... on everything. Some people don't even use the metric system! – user5402 Apr 13 '18 at 13:35
  • @inéquation or even $f(x)|_b^a$ or $[f(x)]_b^a$ or $f(x)]_b^a$. And then we got $n(X)$ or $#X$ or $|X|$ or $\text{card}(X)$. But imho, $]a,b[$ just looks ugly. Like using $\ni$ for "such that". Decimal seperator is a big one, though :) – Mr Pie Apr 13 '18 at 13:41
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    @user477343 I know it's a matter of opinion but let me explain why I prefer $]a,b[$. First the open interval $(a,b)$ and the ordered pair $(a,b)$ shouldnt have the same notation so $x\in (a,b)$ is ambiguous. Second, it helps in geometry. $(AB)$ is a line and $[AB]$ is a line segment. How do you denote the segment $[AB]$ without its endpoints? (my ans: $]AB[$). How do you denote the half line starting at $A$ passing by $B$ and extending indefinitely? ($[AB)$) What if we remove $A$ from this half-line? ( $]AB)$ ). – user5402 Apr 13 '18 at 13:50
  • @inéquation definitely a great point. But (not intending on an argument, but just a civil discussion) is there something better to write than $]...[$? An alternative other than $(...)$? I also denote $AB$ as a line instead of $(AB)$, which would be my notation for a line without its endpoints. – Mr Pie Apr 13 '18 at 13:53
  • @inéquation I would also probably not use math to notate what I mean, and just write it word for word. e.g. we can write something like $$\sum_{p\text{ prime}} p$$ without getting into trouble ... But then there is also Apostol's notation, $$\sum_p ' p$$ – Mr Pie Apr 13 '18 at 14:06
  • $(AB)$ is the line, $AB$ is the distance between $A$ and $B$. Of course we can use words instead of symbols but geometry proofs will double in length this way. I explained why I prefer these notation, no need to push any notation. Just use whichever notation you want. – user5402 Apr 13 '18 at 14:39
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I interpret your question as follows: for any positive integer $n$, there exists a positive integer $z$ such that $z^n$ is a sum of two squares (of integers). But Gauss' theorem on sums of two squares - based on the decomposition of primes in the Gaussian integers - states that a positive integer $z$ is a sum of two squares iff for any prime $p\equiv 3 \bmod 4$, the exponent $v_p(z)$ such that $p^{v_p(z)}$ divides exactly $z$ (this could be $0$) is even . Since $v_p(z^n)=nv_p(z)$, we conclude that : - if $n$ is even, any $z$ will do -if $n$ is odd, $z$ will do iff $z$ itself is a sum of two squares .

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