# Find all non-negative real numbers $a_1$ $\le$ $a_2$ $\le$ $\ldots$ $\le$ $a_n$

Find all non-negative real numbers $$a_1$$ $$\le$$ $$a_2$$ $$\le$$ $$\ldots$$ $$\le$$ $$a_n$$

satisfying

$$\sum_{i=0}^n a_i = 12$$ ,

$$\sum_{i=0}^n (a_i)^2 = 18$$,

and $$\sum_{i=0}^n (a_i)^3 = 27$$

I have a feeling that the answer is not possible, but believe that if we can prove that if all numbers are not rational we can use Fermat's last theorem to show that for the last case we can't have

$$a^n$$ + $$b^n$$ = $$c^n$$ for n is greater than 2

Any help would be appreciated

• $3^2+4^2 = 5^2$. look up the statement of FLT – mathworker21 Mar 5 at 16:00
• We have real numbers, not integers. So $c=\pm \sqrt{a^2+b^2}$ is always possible. – Dietrich Burde Mar 5 at 16:00

\eqalign{\sum_i a_i (1 + t a_i)^2 &= \sum_i a_i + 2 t \sum_i a_i^2 + t^2 \sum_i a_i^3\cr &= 12 + 36 t + 27 t^2 = 3 (2+3t)^2} which is $$0$$ at $$t = -2/3$$. Thus $$\sum_i a_i (1 - 2 a_i/3)^2 = 0$$, which is only possible (for nonnegative $$a_i$$) if all $$a_i = 0$$ or $$3/2$$. We need $$8$$ of the $$a_i$$ to be $$3/2$$ and the others $$0$$.

• Can you explain how you got the sum $\sum_i a_i (1 + t a_i)^2$ please – user101 Mar 5 at 17:45
• A convenient polynomial whose expansion gives you terms in $a_i$, $a_i^2$ and $a_i^3$. – Robert Israel Mar 5 at 21:21

Assume that all those numbers are positive. By Cauchy inequality we have $$(a_1^3+a_2^3+...+a_n^3)(a_1+a_2+...+a_n)\geq (a_1^2+a_2^2+...+a_n^2)^2$$

with equality iff $$a_1^3:a_1 = a_2^3:a_2=...a_n^3:a_n$$ i.e. $$a_1=a_2=...=a_n$$

Since $$27\cdot 12 = 18^2$$ we have eqaulity case here so $$a_i = 12/n$$ and $$a_i^2= 18/n$$ so

$${144\over n^2} = {18\over n}\implies n=8$$

So in general all the sequences are like this: first we have some $$0$$ and last $$8$$ numbers must be $$3/2$$.

• This works if you assume all $a_i > 0$. – Robert Israel Mar 5 at 21:20
• Yes, so then your conclusion should be $n-k = 8$... – Robert Israel Mar 5 at 21:23