UPATE: I asked this question on MO here.
I was solving problem 1.2.52 in "An introduction to the theory of numbers by by Ivan Niven, Herbert S. Zuckerman, Hugh L. Montgomery"
Show that if $\gcd(a, b)$ = $1$ and $p$ is an odd prime, then $\displaystyle \gcd\left(a+b, \frac{a^p+b^p}{a+b}\right)=1$ or $p$.
So I tried to represent $\displaystyle \sum_{k=0} ^m (-1)^ka^{m-k}b^k$ as sum of powers of $a+b$ For any even number $m$ that is one less than a prime number .
$$\sum_{k=0} ^2 (-1)^k a^{2-k}b^k = (a+b)^2 -3ab$$ $$\sum_{k=0} ^4 (-1)^k a^{4-k}b^k = (a+b)^4 -5(ab)(a+b)^2 +5(ab)^2$$ $$\sum_{k=0} ^6 (-1)^k a^{6-k}b^k = (a+b)^6 -7(ab)(a+b)^4 +7 \cdot 2(ab)^2(a+b)^2 -7(ab)^3$$
I noticed that $\displaystyle \sum_{k=0} ^m (-1)^ka^{m-k}b^k=(a+b)^m +(m+1)\left(\sum_{k=1}^{\frac m 2} g_m(k)(-(ab))^k (a+b)^{m-2k}\right)$ where $g_m(k): \mathbb{N} \to \mathbb{Z}$ (I was not able to find a way to determine of $g_m(k)$ )
I couldn't prove this pattern or find $g_m(k )$ or prove that $\displaystyle \sum_{k=0} ^2 (-1)^k a^{2-k}b^k $ can be represented as sum of powers of $a+b$ For any even number $m$ that is one less than a prime number.
So my question is: How to prove that for any $m=p-1$ such that $p$ is prime number $\displaystyle \sum_{k=0} ^m (-1)^ka^{m-k}b^k=(a+b)^m +(m+1)\left(\sum_{k=1}^{\frac m 2} g_m(k)(-(ab))^k (a+b)^{m-2k}\right)$ and determine $g_m(k)$.
I don't think it is possible represent $\displaystyle \sum_{k=0} ^m (-1)^ka^{m-k}b^k$ in the form $\displaystyle(a+b)^m +(m+1)\left(\sum_{k=1}^{\frac m 2} g_m(k)(-(ab))^k (a+b)^{m-2k}\right)$.
I don't think this will work for any even number because I noticed that in any even rows $m$ that is above a prime row $p$ i.e ($p=m+1$) in this image
There is a very interesting property $\dbinom{m}{k}+(-1)^{k+1} |p $ for example $253+1|11 , \ \ 15-1 | 7$ and so on but for any other even row this doesn't hold.