Trouble with a proof. I cannot prove this without inf many proofs for each and every case.

Question

For each and every $k\geqslant 1$, I want to prove whether or not the following equation has infinitely many solutions $(a_1,a_2,b,x)$ over the integers, and on the condition that $a_1\neq a_2\neq b$.

$$\sum_{n=1}^2 a_n^{ \ \ 2} + b^2(k-2) = \left(\sum_{n=0}^k x^n\right)^2.\tag1$$

This is to say, I want to prove that $$a_1^{ \ \ 2} + a_2^{ \ \ 2} + b^2(k - 2) = \big(\underbrace{x^n + x^{n - 1} +\cdots + x^1 + x^0}_{\text{$k$ times}}\big)^2$$ has infinitely many solutions in integers. My attempt, however, is quite exhaustive. I cannot prove that $(1)$ has infinitely many integral solutions for all $k$.

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My Attempt:

I was able to prove that for all $k\in\mathbb{Z}^+$ and $x\in\mathbb{R}$, $$\sum_{n=1}^k nx^{n-1} + \sum_{\substack{n = k \\ m = 1}}^{k-1}x^n(n - 2m + 1) = \left(\sum_{n=0}^k x^n\right)^2,\tag2$$ and it appears that from this equation, I can derive proofs that $(1)$ holds truth for every $k$, but not one entire proof. With the summation that has two indices beneath the sigma but one index above it, to be clear, $$\sum_{\substack{n=k \\ m=1}}^{k-1} x^n(n-2m+1) = \left(\sum_{n=k}^{k-1}x^n\right)\sum_{m=1}^{k-1}(n-2m+1).$$However, for the sake of proving $(1)$ to be true, we consider $x\in\mathbb{Z}$.

The case $k = 1$ is easy. It just shows that $(a_2 + 1)(a_2 - 1) = (b^2 + a_1)(b^2 - a_1)$.

I $k = 2$ I

When we consider $k = 2$, we want to prove that $(*)$ there are infinitely many squares that are the sum of two squares. The following equation is thus produced from $(2)$. That is, we substitute $k = 2$ in $(2)$ and our output is as follows.

$$x^2 + 2x + 1 = (x + 1)^2$$ We can prove the statement $(*)$ from this equation. Simply, either multiply both sides by $(x - 1)^2$ to obtain, $$(x^2 - 1)^2 + (2x)^2 = (x^2 + 1)^2,\tag*{$\because (x-1)^2 = x^2 - 2x + 1$}$$ or let $2x + 1 = u^2$ for some $u\in\mathbb{Z}$ follows that $u = 2v + 1$ for some $v\in\mathbb{Z}$ to obtain, $$(2v^2 + 2v)^2 + (2v + 1)^2 = (2v^2 + 2v + 1)^2.\tag*{$\bigcirc$}$$

I $k = 3$ I

Proof: Consider the equation, $$\begin{align} y &= x^4 + 2x^3 + 3x^2 + 2x + 1 \\ &= (x+1)(x^3 + x^2 + 2x) + 1 \\ &= x(x + 1)(x^2 + x + 2) + 1 \\ &= x^2(x + 1)^2 + 2x(x + 1) + 1 \ \because \ x^2 + x = x(x + 1) \\ &= (x^2 + x + 1)^2.\quad\quad\quad\quad\text{$\big($$(2)$ shows this result anyway.$\big)$}\end{align}$$ Let $3x^2 = x^2 + x^2 + x^2$ follows that, $$\begin{align} (x^2 + x + 1)^2 &= x^4 + 2x^3 + (x^2 + x^2 + x^2) + 2x + 1 \\ &= x^2(x^2 + 2x + 1) + x^2 + (x^2 + 2x + 1) \\ &= x^2 + (x + 1)^2 + x^2(x + 1)^2.\end{align}$$ Therefore, we obtain a desired result for all $x$. $$x^2 + (x+1)^2 + (x^2 + x)^2 = (x^2 + x + 1)^2.\tag*{$\bigcirc$}$$

I $k = 4$ I

Proof: Consider the equation,

$$x^6 + 2x^5 + 3x^4 + 4x^3 + 3x^2 + 2x + 1 = (x^3 + x^2 + x + 1)^2.$$ Let $3x^4 =x^4 + x^4 + x^4$ and $3x^2 = x^2 + x^2 + x^2$ follows that, $$\begin{align} (x^3 + x^2 + x + 1)^2 &= x^6 + 2x^5 + (x^4 + x^4 + x^4) + 4x^3 + (x^2 + x^2 + x^2) + 2x + 1 \\ &= x^4(x^2 + 2x + 1) + (x^2 + 2x + 1) + x^4 + x^4 + 4x^3 + x^2 + x^2 \\ &= x^4(x + 1)^2 + (x + 1)^2 + 2(x^4 + 2x^3 + x^2) \\ &= (x^2)^2(x + 1)^2 + (x + 1)^2 + 2x^2(x^2 + 2x + 1) \\ &= (x + 1)^2 + (x^3 + x^2)^2 + 2x^2(x+1)^2.\end{align}$$ Thus, for all $x$, we obtain as desired. $$(x + 1)^2 + (x^3 + x^2)^2 + 2(x^2 + x)^2 = (x^3 + x^2 + x + 1)^2.\tag*{$\bigcirc$}$$

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I cannot keep proving each case for $k$ ad infinitum (forever and ever). That's impossible! There are infinitely many numbers, and so infinitely many cases! I need to build a proof for the general case $k\in\mathbb{Z}^+$, but I do not know how. Is there a proof of $(1)$ that shows it holds truth for all choices of $k$? If so, please guide me in the right direction.

Thank you in advance.

Answer 1

Take $a_2=0,b=0,a_1=\sum _0 ^k x^{n}$ with x arbitrary. Am I missing something or is this so trivial a question?