The geometrical identity

Proofs

1. By successive divisions

   Performing the successive Euclidian divisions of the polynomial \( (a^n - b^n) \) by \( (a - b) \), we notice that:

   	$$ a-b $$
   $$\hspace{0.6em} a^n - b^n $$	$$ \textcolor{rgb(93 183 129)}{a^{n-1}} $$
   $$ \textcolor{rgb(232 124 124)}{-a^n} \textcolor{rgb(93 183 129)}{+a^{n-1}b} - b^n $$	$$ a^{n-1} \textcolor{rgb(93 183 129)}{+a^{n-2}b} $$
   $$ \hspace{1em} 0 \hspace{0.6em} \textcolor{rgb(232 124 124)}{-a^{n-1}b} \textcolor{rgb(93 183 129)}{+a^{n-2}b^2} - b^n $$	$$ a^{n-1} +a^{n-2}b \textcolor{rgb(93 183 129)}{+a^{n-3}b^2} $$
   $$ \hspace{4.2em} 0 \hspace{1.2em} \textcolor{rgb(232 124 124)}{- a^{n-2}b^2} \textcolor{rgb(93 183 129)}{+ \dots + a^{n-p}b^{p} + \dots} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 \textcolor{rgb(93 183 129)}{+ \dots + a^{n-p-1} b^{p}} $$
   $$ \hspace{8.2em} 0 \hspace{1.6em} + \dots \textcolor{rgb(232 124 124)}{-a^{n-p}b^{p}} \textcolor{rgb(93 183 129)}{+ a^{n-p+1}b^{p+1} } + \dots - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} \textcolor{rgb(93 183 129)}{ + a^{n-p} b^{p+1}} $$
   $$ \hspace{15.2em} 0 \hspace{1.4em} \textcolor{rgb(232 124 124)}{- a^{n-p+1}b^{p+1}} + \dots - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p+1} $$
   $$ \hspace{20.4em} \dots \dots \text{until } (p = n - 3)$$	$$ $$
   $$ \hspace{20.4em} 0 \hspace{2em} \textcolor{rgb(93 183 129)}{ + a^2 b^{n-2}} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p+1} + \dots \textcolor{rgb(93 183 129)}{+ ab^{n-2} } $$
   $$ \hspace{23em} \textcolor{rgb(232 124 124)}{-a^2 b^{n-2}} \textcolor{rgb(93 183 129)}{+ a b^{n-1}} - b^n $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2 + \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} \textcolor{rgb(93 183 129)}{+ b^{n-1} } $$
   $$ \hspace{27.4em} \textcolor{rgb(232 124 124)}{- a b^{n-1}} \textcolor{rgb(232 124 124)}{+ b^n } $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} $$
   $$ \hspace{30em} 0 $$	$$ a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} $$

   Therefore we finally obtain,

   $$a^n - b^n = (a-b) \Bigl[a^{n-1} + a^{n-2}b +a^{n-3}b^2+ \dots + a^{n-p-1} b^{p} + a^{n-p} b^{p-1} + \dots + ab^{n-2} + b^{n-1} \Bigr] $$

   And as a result of it,

   $$\forall n \in \mathbb{N}^*, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{R}^2,$$
   $$ a^n - b^n = (a-b) \sum_{p=0}^{n-1} a^{n-p-1}b^p \qquad \text{(Geometrical identity)} $$

2. By induction

   We want to prove the following proposition \((P_n)\) by mathematical induction:

   $$\forall n \in \mathbb{N}^*, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{R}^2,$$
   $$a^n - b^n = (a-b) \sum_{p=0}^{n-1} a^{n-p-1}b^p \qquad (P_n)$$

   1. First term calculation

      Let's check if the proposition holds true for the first term, i.e., when \( n = 1 \).

      $$a^1 - b^1 = a-b $$

      On the other hand, the right-hand side yields:

      $$ (a-b) \sum_{p=0}^{1-1} a^{1-p-1}b^p $$
      $$ (a-b) \sum_{p=0}^{0} a^{0}b^0 $$
      $$ = a-b $$

      Therefore, we successfully verify that:

      $$ a^1 - b^1 = (a-b) \sum_{p=0}^{1-1} a^{1-p-1}b^p = a - b$$

      \((P_1)\) is true.

   2. Inductive step

      Let \( k \in \mathbb{N}^* \) be a non-zero natural integer.

      We assume that the proposition \((P_k)\) is true for this rank \( k \):

      $$ a^k - b^k = (a-b) \sum_{p=0}^{k-1} a^{k-p-1}b^p \qquad (P_k) $$

      Let's check if the property holds for the next rank \( (k + 1) \):

      $$ a^{k+1} - b^{k+1} = (a-b) \sum_{p=0}^{k} a^{k-p}b^p \qquad (P_{k + 1}) $$
      
      

      Let's start from the left-hand side of our target expression and rearrange the terms:

      $$ a^{k+1} - b^{k+1} = a \cdot a^k - b \cdot b^k $$

      Using a classic algebraic trick, let's insert the cross-term \( -a \cdot b^k + a \cdot b^k \):

      $$ a^{k+1} - b^{k+1} = a \cdot a^k - a \cdot b^k + a \cdot b^k - b \cdot b^k $$

      Next, we factor out \(a\) and \(b^k\) respectively:

      $$ a^{k+1} - b^{k+1} = a(\underbrace{a^k - b^k} _{(P_k)}) + b^k(a - b) $$

      By substituting our induction hypothesis \( (P_k) \), we get:

      $$ a^{k+1} - b^{k+1} = a \cdot \left[ (a-b) \sum_{p=0}^{k-1} a^{k-p-1}b^p \right] + b^k(a - b) $$

      Finally, we factor out the common block \((a-b)\):

      $$ a^{k+1} - b^{k+1} = (a-b) \left[ \sum_{p=0}^{k-1} a^{k-p}b^p + b^k \right] $$

      The isolated term \( b^k \) perfectly fits into the summation as the term for index \( (p = k) \):

      $$ a^{k+1} - b^{k+1} = (a-b) \left[ \sum_{p=0}^{k-1} a^{k-p}b^p + \underbrace{\sum_{p=k}^{k} a^{k-p}b^p} _{b^k} \right] $$
      $$ a^{k+1} - b^{k+1} = (a-b) \sum_{p=0}^{k} a^{(k+1)-p-1}b^p $$

      Which simplifies to our expected formula, confirming that \((P_{k + 1})\) is true:

      $$ a^{k+1} - b^{k+1} = (a-b) \sum_{p=0}^{k} a^{k-p}b^p \qquad (P_{k + 1}) $$

   3. Conclusion

      The proposition \((P_n)\) is true for its base case \(n_0 = 1\) and is hereditary for any \(k \in \mathbb{N}^*\).

      By the principle of mathematical induction, it is true for all \(n \in \mathbb{N}^*\).

      Thus, we successfully establish that:

      $$\forall n \in \mathbb{N}^*, \enspace \forall (a, b) \in \hspace{0.04em} \mathbb{R}^2,$$
      $$ a^n - b^n = (a-b) \sum_{p=0}^{n-1} a^{n-p-1}b^p \qquad \text{(Geometrical identity)} $$