Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$

Question

Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.

I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers will, when plotted in 3D space, generate a hypotenuse whose length is also an integer. For example; $\sqrt {6^2 + 42^2 + 85^2} = 95$

What I've found is quite interesting. It seems that there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x=2^y$ or $x=5 \times 2^y$ and $y$ is an integer.

$$\begin{array}{c|c|c} y & 2^y & 5 \times 2^y \\ \hline 0 & 1 & 5 \\ \hline 1 & 2 & 10 \\ \hline 2 & 4 & 20 \\ \hline 3 & 8 & 40 \\ \hline 4 & 16 & 80 \\ \hline 5 & 32 & 160 \\ \hline 6 & 64 & \text{see below} \\ \hline 7 & 128 & \text{see below} \\ \end{array}$$

So, there are no solutions for $\sqrt {a^2 + b^2 + c^2} = x$ when $x$ equals any value from the following series: $1,2,4,5,8,10,16,20,32,40,64,80,128,160...$

I wrote a program which brute-force checked this for values up to $250$ for $a, b, \text {and } c$ (which is why I show no solutions to $5 \times 2^x$ using $6$ and $7$ above, since they should be $320$ and $640$ respectively). The program found at least 1 solution for all other integers from $1 \text{ to }250$.

Here's the code I used (excel VBA): http://pastebin.com/jVA27jYp

Answer 1

It appears you do not permit any of $a,b,c$ to be zero. You can write it as $a^2+b^2+c^2=x^2$. In that case, it comes from the fact that the squares $\bmod 8$ are $0,1,4$ To have $2^{2y}$ be the sum of three squares you would have to have all of the squares be $0 \bmod 8$ or one be $0 \bmod 8$ and two of them be $4 \bmod 8$. In either case you can divide $a,b,c$ by $2$ and get a solution for $2^{2y-2}$, but then you make the same argument and there is no soluton by infinite descent. Similarly, if $x^2=5^22^{2y}$ you can perform the same infinite descent to get to $x^2=25$ and you have checked the cases for that.

Answer 2

WLOG, $a,b,c,y$ are positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ then for some $y$ then there is a solution with a least value of $\min (a,b,c).$ Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contradicts the minimality of $\max (a,b,c)$, unless $y=1$.But there is no solution with $y=1.$

But if $a,b,c$ are not all even then $a^2+b^2+c^2$ is congruent to $1,2,$ or $3$ (modulo $4$) so it cannot be divisible by $4,$ so it can't belong to $\{4^y,25\cdot 4^y\}.$

Answer 3

There is no solution with $y= 0.$

Let $a,b,c,y$ be positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ for some $(a,b,c,y)$ then there is a solution with a least $y$. Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contradicts the minimality of $y$, unless $y=1$.But $y=1$ and $a,b,c$ all even implies there is a solution $(a/2,b/2,c/2,y')$ with $y'=0.$

But if $a,b,c$ are not all even then $a^2+b^2+c^2$ is congruent to $1,2,$ or $3$ (modulo $4$) so it cannot be divisible by $4,$ so it can't belong to $\{4^y,25\cdot 4^y\}.$