Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How would I go about proving the following?

If $a$, $b$, $c$, $n$ are positive integers, then

$a^2+b^2+c^2 \neq 2^nabc$

I tried doing something similar to the proof for Adrien-Marie Legendre's Three Square theorem: $a^2+b^2+c^2=n$ iff there are not integers $k$, and $m$ so that $n=4^k(8m+7)$. It didn't quite work out...

$2^nabc$ is always even. So if $a^2+b^2+c^2 = 2^nabc$, then $a^2+b^2+c^2$ must be even.

That means there is $a_1$, $b_1$, $c_1$ so that $a = 2a_1$, $b = 2b_1$, and $c = 2c_1$

So $(2a_1)^2+(2b_1)^2+(2c_1)^2 = 2^nabc \rightarrow 2(2a_1^2+2b_1^2+2c_1^2)= 2^nabc$

and we get $2a_1^2+2b_1^2+2c_1^2= 2^{n-1}abc$

We can continue to do this procedure with $a_2$, $b_2$, $c_2$ then $a_3$, $b_3$, $c_3$ then ... $a_n$, $b_n$, $c_n$.

With $a_n$, $b_n$, $c_n$ we'd get

$2^na_n^2+2^nb_n^2+2^nc_n^2= 2^{n-n}abc=abc$

Since $a_n=2a_{n-1}$ and $a_0=a$,

$a_n = \frac{a}{2^n}$

and we get


This just becomes the original equation. $a^2+b^2+c^2 = 2^nabc$

share|cite|improve this question
I suppose I could just use Legendre's theorem to complete the proof: $2^nabc$ is the sum of three squares iff there are integers $k$ and $m$ so that $4^k(8m+7)=2^nabc$ We can pull out a power of 4: $2^nabc=2^{\frac{n}{2}}2^{\frac{n}{2}}abc=4^\frac{n}{2}abc$, but $abc$ cannot be represented as $8m+7$ because $abc$ is even and $8m+7$ is odd. – objectivesea Nov 2 '11 at 8:24
You have the theorem reversed: a positive integer is a sum of three squares if and only if it is not of the form $4^n (8m+7).$ – user7530 Nov 2 '11 at 8:32
up vote 5 down vote accepted

Since the right-hand side is even, either exactly one or all three of $a,b,c$ must be even.

The former case is impossible, as you can easily see by taking both sides mod 4.

In the latter case, let $2^k$ be the greatest power of 2 in the GCD of $a,b,c$. Then

$$\left(\frac{a}{2^k}\right)^2 + \left(\frac{b}{2^k}\right)^2 + \left(\frac{c}{2^k}\right)^2 = 2^{n+k} \frac{a}{2^k} \frac{b}{2^k} \frac{c}{2^k},$$ with at least one of the terms on the left-hand side odd, and we are back in case 1.

EDIT: Note that $n>0$ is essential. When $n=0$, $$3^2 + 3^2 + 6^2 = 54 = 3\cdot3\cdot6.$$

share|cite|improve this answer
@Phira $n$ is a positive integer. – user7530 Nov 2 '11 at 14:23
I'm trying to figure out why the final step is permissible. $2^{n+k}\frac{a}{2^k}\frac{b}{2^k}\frac{c}{2^k}=2^nabc$, but $(\frac{a}{2^k})^2+(\frac{b}{2^k})^2 +(\frac{c}{2^k})^2 \neq a^2+b^2+c^2$ unless $k=0$. – objectivesea Nov 2 '11 at 15:44
Since each of $a, b, $ and $c$ are divisible by $2^k$, both sides of the equation must be divisible by $2^{2k}$. Carrying out this division gives the equation in my answer. The right-hand side in your comment above is incorrect: it should be $2^{n-2k}abc$. – user7530 Nov 2 '11 at 15:57
@Phira No, you can't include this case, as the statement is false when $n=0$. I've added a counterexample to my answer. – user7530 Nov 2 '11 at 16:12
Thanks. Will delete my comments. – Phira Nov 2 '11 at 18:21

To go from $a^2+b^2+c^2$ even to $a,b,c$ even in this case, you need an argument, although it is true for $n$ strictly positive.

You should try to express your equation for only $a_k$, $b_k$ and $c_k$, then you will see that your argument does not end after $n$ steps. Either, you argue with infinite descent or equivalently, you divide immediately by the largest factor of 2 possible.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.