Why does $n*T_n = m*T_m$ imply $T_\text{n+m} = 0$?

Question

I was fiddling around with Arithmetic Progressions and I noticed this pattern.

\begin{align} n*T_n = m*T_m \implies T_\text{n+m} = 0 \end{align}

where $n, m \in \{0, \mathbb{Z}^{+}\}$ and $n \neq m$.

I could prove that it's true like so:-

\begin{align*} & m*T_m = n*T_n \\ & \Rightarrow m\{a + (m-1)d\} = n\{a + (n-1)d\} \\ & \Rightarrow ma + m^2d - md = na + n^2d - nd \\ & \Rightarrow am - an + m^2d - n^2d - md + nd = 0 \\ & \Rightarrow a(m-n) + (m^2-n^2)d - (m-n)d = 0 \\ & \Rightarrow a(m-n) + (m-n)(m+n)d - (m-n)d = 0 \\ & \Rightarrow (m-n)\{a + (m+n)d - d\} = 0 \\ & \Rightarrow (m-n)\{a + (m+n-1)d\} = 0 \\ & \Rightarrow a + (m+n-1)d = 0 \\ & \Rightarrow T_{m+n} = 0 \end{align*}

But I couldn't figure out the intuitive reasoning behind it. Could you help me out using as little machinery as possible?

Answer 1

An intuitive explanation with up to middle school level analysis: First we rewrite $T_n = T_1 + (n-1)a$ as $T(x) = ax + b$, where $b = T_1$, and $x = n - 1$.

$x(T(x)) = ax^2 + bx = Q(x)$, which is a quadratic expression. And, if for some $m,n \in \mathbb{N}$, $nT(n) = mT(m)$, then we can think of this as saying: $$an^2 + bn = am^2 + bm \\Q(n) = Q(m)$$ We also want to say that if above is true, then $m+n = \frac{-b}{a}$, since that is the root to our linear function $T(x)$. However, you may notice that $\frac{-b}{a}$ is also a root to our quadratic expression $Q(x) = ax^2 +bx = x(ax + b)$. Hence (since $m, n \in \mathbb{N}$ and $m \not = n$, $ m + n \not = 0$ so) if $Q(m+n) = 0$, then $T(m+n) = 0$.

So our final question becomes, why is it that if $Q(m) = Q(n)$, then $Q(m+n) = 0$? Because we know our apex is at $\frac{-b}{2a}$ and since $m$ and $n$ should be on opposite side of it, we can represent $m = \frac{-b}{2a} + d$ and $n = \frac{-b}{2a} - d$, WLOG. Then, $$Q(m+n)= Q(\frac{-b}{2a} + d + \frac{-b}{2a} -d) = Q(\frac{-2b}{2a}) = Q(\frac{-b}{a}) = 0$$ And since $Q(m+n) = 0$, $T(m+n) = 0$. $\square$

Im not sure if this is as intuitive as you liked, but this is about as good of a picture as I can paint for you. Hopefully this helps, let me know if I can fix it in any way!

Answer 2

Disclaimer: This essentially uses Vieta's formula, which arguably is more machinery than "algebraic manipulation". Though to be fair, almost any other method will be more machinery than expand and check, and this approach yields some nice results.

1. Think of $T_n$ as just a linear equation $T_n = (a-d) + nd$.
   • That's all that really matters here.
   • $ d \neq 0$ by assumption.
2. $n \times T_n = m \times T_m$ for $ n \neq m$ means that $n,m$ are the distinct roots to the quadratic $ dx^2 + (a-d) x = k$ for some $k$.
   • $ d \neq 0 $ means we have a quadratic.
3. Lemma: If $f(x)$ is a polynomial, then the sum of the (possibly complex, with multiplicity) solutions to $f(x) = K$ is a constant for any constant $K$.
   • Proof is left to the reader. Look up Vieta's formula if you're stuck.
4. Corollary: Let $ f(x) = dx^2 + (a-d) x$, then the sum of solutions to $f(x) = K$ is $n+m$.
5. Corollary: Apply this to $K = 0$, since $ x= 0 $ is clearly a root to $f(x) = 0$, so the other root is $n+m - 0 $.
   • Thus, $ (n+m) T_{n+m} = 0 \times T_0 = 0$.
6. Bonus Corollary: $nT_n = mT_m \Leftrightarrow p T_p = (n+m - p) T_{n+m - p } $ for all $p$.
   • Your result is the special case of $p = 0 $.
7. Bonus Corollary: $nT_n = mT_m$ iff $\frac{a-d}{d} = n+m$.
8. Bonus Corollary: At no point did we require that $n, m$ are integers. So this also holds for (say) $\frac{1}{2} T_{\frac{1}{2} } = \pi T_\pi$, even though there isn't a corresponding arithmetic progression term.
   • Which is why I said at the start to just think of $T_n$ as a linear equation.

Answer 3

From the property of AP or linear functions, as worded by @preferred_anon,

$$\begin{align*} \frac{T_{n+m} - T_n}{m} &= \frac{T_{n+m} - T_m}{n} &&(=d)\tag{*}\\ nT_{n+m} - nT_n &= mT_{n+m} - mT_m\\ nT_{n+m} &= mT_{n+m} \end{align*}$$

This means that either $T_{n+m} = 0$ or $m=n$ (rejected).

(The $(*)$ fractions are undefined if either of $m,n$ is zero, but the next line still holds.)

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Or as I initially presented it,

$$\begin{align*} T_{n+m} - T_n &= T_m - T_0\\ T_{n+m} &= T_n + \left(T_m - T_0\right)\tag1\\ n T_{n+m} &= nT_n + n\left(T_m - T_0\right)\\ &= mT_m + n\cdot md\\ &= mT_m + m\left(T_n - T_0\right)\\ &= m\left(T_m + T_n - T_0\right)\\ &= m T_{n+m}&&\text{using line $(1)$} \end{align*}$$

This means that either $T_{n+m} = 0$ or $m=n$ (rejected).

Answer 4

Notice that, at no point during the proof, you used the fact that your coefficients were integers. In fact, you can deduce this from a simple geometric argument and calculus.

Consider the function $f(x) = x(a(x-1)+d)$, whose graph is a parabola. If you find two different points $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$, then their midpoint $\frac{x_1+x_2}{2}$ will necessarily be where $f$ attains its extremum. In other words, $f'(\frac{x_1+x_2}{2}) = 0$, and so $a(x_1+x_2 - 1) + d$.