For any natural number $n$, exists $z,x,y\in\mathbb{N}$ such that: $z^n=x^2+y^2$

Question

Prove that for any natural number $n$, one can find $z,x,y\in\mathbb{N}$ such that: $z^n=x^2+y^2.$

Here $\mathbb{N}$ denotes all the natural numbers $n\geq1$.

This problem feels like so much general that i don't even know how to start. I don't think that any of the things that i tought is worthy of posting here.

So i'm asking for hints. Can someone give some good ideas on how to start?

pythagorean theorem might be useful to prove or disprove cases if n is in the natural numbers. — , Sep 13 '17 at 22:35
Prove that the product of two numbers which can be represented as a sum of two squares can also be represented as the sum of two squares; then it's straightforward. There is also an even easier thing you can do if $n$ is odd, for starters. — Qiaochu Yuan, Sep 13 '17 at 22:40
@QiaochuYuan But not, if $0$ is not allowed. $2=1^2+1^2$, but $4$ has no representation. — Peter, Sep 13 '17 at 22:41
That only rules out a very small number of cases; if you start with $5 = 1^2 + 2^2$ there won't be a problem. Although Wen's solution below is even cleaner. — Qiaochu Yuan, Sep 13 '17 at 22:42
Do you know about $\mathbb Z[i]$? In general, $(u+vi)^n=x+y i$ then $x^2+y^2=(u^2+v^2)^n$. — Thomas Andrews, Sep 13 '17 at 22:55
This gives most answers, and if $u\neq \pm v$ and $uv\neq 0$ then you can get $x,y,z\in\mathbb N$. — Thomas Andrews, Sep 13 '17 at 22:56
Alternative notation that is actually preferred: $(x,y,z)\in\Bbb N^3$ — gen-ℤ ready to perish, Sep 13 '17 at 23:37
https://artofproblemsolving.com/community/c3046h1124891_almost_pythagoras — individ, Sep 14 '17 at 04:51

score 8 · Answer 1 · edited Sep 13 '17 at 22:51

8

For $n$ even take $3\times 5^{\frac{n-2}{2}},4\times 5^{\frac{n-2}{2}},(5\times 5^{n-1})^\frac{1}{n}=5$

For $n$ odd find $k$ such that $n|2k+1$. Then take $2^k,2^k,2^{\frac{2k+1}{n}}$.

edited Sep 13 '17 at 22:51

user84413

27,211

answered Sep 13 '17 at 22:40

Wen

1,966
9
18

Peter · Answer 2 · 2017-09-13T23:03:19.907

4

Begin with $$5=2^2+1^2$$

If $m=a^2+b^2$ with positive integers $a,b$, then $5m=5a^2+5b^2=(2b+a)^2+(2a-b)^2$

WLOG , we have $a\ge b$, so $2a-b$ is positive, so we have a representation again.

So, with induction, we can easily show that for every $5^n$, there is a solution.

edited Sep 13 '17 at 23:03

answered Sep 13 '17 at 22:37

Peter

84,454

score 4 · Answer 3 · answered Sep 13 '17 at 22:57

Idea: use complex numbers.

Write $$(a+i b)^ n = x + iy$$

Then we get for the norms $$(a^2 + b^2)^n = x^2 + y^2$$

Example: for $n=3$ $$(a+ib)^3 = a^3 - 3 a b^2 + i ( 3 a^2b -b^3)= x+ i y$$ gives the identity $$(a^2 + b^2)^3 = (a^3 - 3 a b^2)^2 + (3 a^2 b - b^3)^2$$

You can do this for every $n$.

score 2 · Accepted Answer · answered Sep 14 '17 at 04:22

2

S is the set of numbers in form $x^2+y^2$ where $x,y\in\mathbb N$. Then

1) If $a,b\in S$ then $ab\in S$. Indeed, $a=x^2+y^2$ and $b=z^2+t^2$ then $ab=(xz+yt)^2+(xt-yz)^2$.

2) $2=1^2+1^2$, $5=1^2+2^2$, ... then $2^n,5^n, ...\in S$.

These answers your question.

answered Sep 14 '17 at 04:22

Jie Fan

1,886

I think the case that $a,b\in S$ implies $ab\in S$ must be investigated further. Because $x,y,z \in \mathbb{N}$, care must be taken that $xt-yz \ne 0$ as $0 \notin \mathbb{N}$ and hence $ab\notin S$. As a counter-example, $x^2 + y^2 = 2^2$ has no solutions in $\mathbb{N}$, but according to your solution, $2^n \in S \ \forall n \in \mathbb{N}$ – thornsword May 30 '20 at 04:14
As an alternate solution, $(x,y,z) = (3.5^{n-1}, 4.5^{n-1}, 5^2)$ works for all $n \in \mathbb{N}$ – thornsword May 30 '20 at 04:16

score 1 · Answer 5 · answered Sep 14 '17 at 00:01

$44^2 + 9^2 = 2017.$ At each stage, with $a^2 + b^2 = 2017^n,$ take $$ (a,b) \mapsto (44a - 9 b, \; \; 9a + 44b) $$ to get $2017^{n+1}.$ This will never give either zero because $\arctan \frac{44}{9}$ is not a rational multiple of $\pi.$ All we are doing is complex numbers, $$ (44 + 9i)(a+bi) = (44a - 9b) + (9a+44b)i \; . $$

Sometimes one or both entires is negative, just take absolute value for that one to get your statement for natural numbers.

44   9  2017 =  2017
1855   792  4068289 =  2017^2
74492   51543  8205738913 =  2017^3
2813761   2938320  16550975387521 =  2017^4
97360604   154609929  33383317356629857 =  2017^5
2892377215   7679082312  67334151108322421569 =  2017^6
58152856652   363911016663  135812982785486324304673 =  2017^7
-716473457279   16535460443040  273934786278325916122525441 =  2017^8
-180343976107636   721111998378249  552526463923383372819133814497 =  2017^9
-14425142934140225   30105832143674232  1114445877733464262976192903840449 =  2017^10
-905658778395237988   1194830327914404183  2247837335388397418422981087046185633 =  2017^11
-50602459200620109119   44421605422676642160  4533887905478397592959152852572156421761 =  2017^12
-2626302653631374580676   1499128505792191272969  9144851905349927944998611303638039502691937 =  2017^13

For any natural number $n$, exists $z,x,y\in\mathbb{N}$ such that: $z^n=x^2+y^2$

5 Answers5