Can every statement be solved by mathematical induction ? (see details below)

Question

I have the following equation system :

$$\sum_{i=1}^n a_i^2 = n $$ $$\sum_{i=1}^n a_i=n$$

here the solution is only $a_i$ =1 .

Can it be solved by mathematical induction ? I have tried , but have not been able to. And I think it's not possible to solve this system of equation by induction. But can anyone prove it ,that indeed this equation system can't be solved by mathematical induction ?

and what are the notions of solvability by induction (if there be any ) ?(i mean, is there any such rule or notion by which it can be determined which statements are solvable by induction and which are not ?

Answer 1

There is no hard-and-fast rule that tells you which statements are provable by induction and which are not, although many statements that depend on the arbitrary choice of a (positive) integer are viable for induction.

The only test you can really perform is to check if you can reduce the statement for $n$ to the statement for $n-1$, which is half of the proof by induction already...

You should note, though, that many of this kind of equations are better solved by a recursive approach: instead of "counting up" to infinity you "count down" to the base case. This works best when your equation has a self-similar structure, in the sense that it can be reduced to a smaller equation of the same type. You can find an example here.

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My original answer follows. Note that before it was edited, the summations in the question were indexed from $0$ to $n$.

I don't think that you can all the solutions of this system of equations (directly) by induction, because I can't think of a way to reduce the equations of the case for $n$ to the equations of the case for $n-1$.

On the other hand, you can solve this geometrically, and in what follows I'm going to assume that you are looking for solutions with $a_0,\dotsc,a_n \in \Bbb{R}$. Now, observe that for any $k \in \Bbb{R} \setminus \{0\}$ the equation $$ a_0^2 + \dotsb + a_n^2 = k $$ defines the (hyper)sphere $S \subset \Bbb{R}^{n+1}$ of radius $\sqrt{k}$ and centre $O = (0,\dotsc,0)$, while the equation $$ a_0 + \dotsb + a_n = k $$ defines the (hyper)plane $\pi \subset \Bbb{R}^{n+1}$ through the $n+1$ points $(k,0,\dotsc,0),(0,k,0,\dotsc,0),\dotsc,(0,\dotsc,0,k)$. Furthermore, note that the closest point of $\pi$ to $O$ is the point $p$ of $\pi$ with $a_0 = \dotsc = a_n = k/(n+1)$.

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Now, from what you say in the rest of your question I believe that you are actually trying to solve the system $$ \begin{cases} a_1^2 + \dotsc + a_n^2 = n \\ a_1 + \dotsc + a_n = n \end{cases} $$ which is the same as setting $k = n+1$ in the above discussion. Then all you have to do is note that (by construction) the distance between any point of $S$ and $O$ is exactly $\sqrt{n+1}$ while the distance between any point of $\pi$ and $O$ is at least $\|p\| = \sqrt{n+1}$, with equality achieved only at $p$. Thus in this case the only solution is indeed $p = (1,\dotsc,1)$.

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What about the system $$ \begin{cases} a_0^2 + \dotsc + a_n^2 = n \\ a_0 + \dotsc + a_n = n \end{cases} $$ though? Since in this case $\|p\| = \sqrt{\frac{n}{n+1}} < \sqrt{n}$ it follows that set of solutions in $\Bbb{R}^{n+1}$ is a circle. For example, for $n = 2$ we have (relabelling $a_0,a_1,a_2$ to $x,y,z$) $$ \begin{cases} x^2 + y^2 + z^2 = 2 \\ x + y + z = 2 \end{cases} \leftrightarrow \begin{cases} x^2 + y^2 + (2 - x - y)^2 = 2 \\ z = 2 - x - y \end{cases} $$ and its projection on the $xy$-plane is

Finally, note that the integer solutions of this system are the $n$ points with $(0,1,\dotsc,1),(1,0,1,\dotsc,1),\dotsc,(1,\dotsc,1,0)$. Indeed, the first equation implies that at least one coordinate must be zero, but as soon as one coordinate is $0$ we are in the previous case.

Answer 2

Your solution as given is not correct: the correct solution is

$$a_0=0,\quad a_n=1 \text{ for }n>0$$

Here is one proof that uses only your second equation in the system. It also uses the inductive definition of the sum symbol:

$$\sum_{i=r}^r a_i=a_r$$ $$\sum_{i=r}^s a_i=\sum_{i=r}^{s-1} a_i + a_s \quad\text{for }s>r$$

If this is not your definition of the sum symbol, you can probably use induction with your definition to prove the above.

Given these, and assuming that $\sum_{i=0}^n a_i = n$,

$$a_0=\sum_{i=0}^0 a_0=0$$

so $a_0=0$. Now assume that $n>0$.

$$a_n=\sum_{i=0}^n a_0-\sum_{i=0}^{n-1} a_0=n-(n-1)=1$$

The first equation in your system is not needed for this proof, but it is also true. That follows easily from $a_i^2=a_i$ since $a_i=0$ or $a_i=1$.

This could be considered a proof by induction, by first proving the inductive definition of the sum symbol. Does this meet your needs?