By Legendre's theorem $\exists a,b,c$ such that $\sum a^2=3^{2^n} \cdot \gcd(a,b,c)^2$ let $a= \gcd(a,b,c) \cdot x...$ then $\sum x^2=3^{2^n}$ and $\gcd (x,y,z)=1$

I can't find the theorem...

Gruberr69 wrote: I can't find the theorem... Well that's not the theorem itself. The real theorem states $n=a^2+b^2+c^2$ $\iff$ $n \neq 4^k(8m+7)$. If $4^k(8m+7)=3^{2^n} \cdot \gcd(a,b,c)^2$ then $8m+7$ is perfect square ie $x^2 \equiv -1 (mod 8)$ which is not true so we can write $\sum a^2=3^{2^n} \cdot \gcd(a,b,c)^2$ As for the original theorem, take a look at https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem

SidVicious wrote: By Legendre's theorem $\exists a,b,c$ such that $\sum a^2=3^{2^n} \cdot \gcd(a,b,c)^2$ let $a= \gcd(a,b,c) \cdot x...$ then $\sum x^2=3^{2^n}$ and $\gcd (x,y,z)=1$ Eh Although$ 3^{2^n} \cdot \gcd(a,b,c)^2$ can be represented as the sum of three perfect square numbers but how can you claim that the three perfect square numbers are just $a^2. b^2$ and $c^2$......

Ok, thanks Stranger8 for correcting me! I will use: $(x^2+y^2+z^2)^2=(x^2+y^2-z^2)^2+(2xz)^2+(2yz)^2$ Notice that $3^2=1^2+2^2+2^2$ then $3^4=(1^2+2^2+2^2)^2=(2^2+2^2-1^2)^2+(2\cdot 1\cdot 2)^2+(2\cdot 1\cdot 2)^2=7^2+4^2+4^2$ then similarly $3^8=(7^2+4^2+4^2)^2=17^2+56^2+56^2$. Now induction: Base is ok. Now suppose $3^{2^n}=2x^2+y^2$ (such numbers exists and we get them by using the above algorithm) with $gcd(x,x,y)=gcd(x,y)=1$. Then $3^{2^{n+1}}=(2x^2-y^2)^2+2(2xy)^2$ suppose there is $p$ such that $p|2x^2-y^2,2xy$. Obviously $p \neq 2$ (cause $3^{2^{n+1}}$ is odd)so we have $p|2x^2-y^2,xy$ From $p|xy$ we have 2 cases: $1.$ $p|x$ then from $p|2x^2-y^2$ we have $p|y$ thus $p|x,y$ not true because $gcd(x,y)=1$. $2.$ $p|y$ then from $p|2x^2-y^2$ we have $p|2x^2$ thus $p|x^2$ thus $p|x$ so again $p|x,y$ not true cause $gcd(x,y)=1$. So from this $p=1$ and now $gcd(2x^2-y^2,xy,xy)=1$ and $3^{2^{n+1}}=(2x^2-y^2)^2+2(2xy)^2$ and we are done

Nice solution

My solution :notice that if $gcd(x,y,z)$ is not 1,then $gcd(x,y,z)|3$ ,it tells us that $gcd(x,y,z)=3$,so we want one of x,y,z is not divisible by 3,we go by induction ,we have $(2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2=9(a^2+b^2+c^2)$ if $3|(2a+2b-c)$ then we change $b$ into $-b$ (b is not divisible by 3) applying this for $2^{n-1}$times and we are done And my solution also shows that this problem is true for $3^n$

Another solution: Just show that $3^{2^n}-1$ is a sum of two squares, which is easy. $3^{2^n}-1=(3^{2^{n-1}}-1)(3^{2^{n-1}}+1)=(3^{2^{n-2}}-1)(3^{2^{n-2}}+1)(3^{2^{n-1}}+1)=\cdots = 2\prod_{k=0}^{n-1}(3^{2^k}+1)$ All of the factors of the product are clearly sums of two squares. So the whole product is a sum of $2$ squares by the Diophantus identity: $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$