Prime factor of $A=14^7+14^2+1$ Find a prime factor of $A=14^7+14^2+1$. Obviously without just computing it.

[Update: see the method of simpler multiples for a more general perspective]
It's a special case of $\ \bbox[5px,border:1px solid #c00]{x^{2}\!+\!x\!+\!1\mid x^A\! +\! x^B\! +\! x^C\ \ {\rm if} \ \ \{A,B,C\}\equiv \{2,1,0\}\pmod{\!3}}$
Hint $\ $ Notice: $\ {\rm mod}\,\  x^2\!+\!x\!+\!1\!:\ \color{#c00}{x^3\equiv 1}\,\Rightarrow\ x(\color{#c00}{x^6})\!+\!x^2\!+\!1\equiv x\!+\!x^2\!+\!1\equiv 0$
Remark $ $ i.e. replacing $\,\color{#0a0}x\,$ by $\,x^7\,$ in $\, f = x^2+\color{#0a0}x+1 = (\color{#c00}{x^3\!-\!1})/(x\!-\!1) $ it remains divisible by $f$ because $\,{\rm mod}\ f\!:\ \color{#c00}{x^3\equiv 1}\,\Rightarrow\, x^7 \equiv\, x(\color{#c00}{x^3})^2\equiv\color{#0a0} x$
More generally if we replace $\,x^{\large k}\,$ by $\, x^{\large k+jn}\,$ in $\, f = (\color{#c00}{x^n\!-1})/(x\!-\!1)\,$ it remains divisible by $f$ since $\,{\rm mod}\ f\!:\ \color{#c00}{x^n\equiv 1}\,\Rightarrow\, x^{\large k+jn} = x^{\large  k}(\color{#c00}{x^{\large  n}})^{\large  j}\equiv x^k$
This viewpoint naturally leads to the following
Theorem $ $ If $\ f = \sum_k f_k x^k\,$ divides $\,\color{#c00}{x^n-1}$ then $\,f\,$ divides $\, \sum_k f_k x^{\large h_k}\,$ if $\,h_k \equiv k\pmod{\! n}$
Proof $\ $  As above $\!\bmod f\!:\ x^{\large h_k}\! = x^{\large k+jn}\equiv x^k\ $  so $\ \sum_k f_k x^{\large h_k}\equiv \sum_k f_kx^k\equiv f\equiv 0.\ \ $ QED
For example $\ \bbox[5px,border:1px solid #c00]{x^2\!+\!x\!+\!1 \mid x^{\large 2+3i}\!+x^{\large 1+3j}\!+x^{\large 3k}}\ $ generalizes the OP $ $ (where $\,i,j,k = 0,2,0)$
This method often helps one to recognize multiples of (cyclotomic) factors of $\,x^n-1\,$ having such "tweaked terms". Similarly for factors of binomials or other "few monomial" polynomials. Note also there are efficient algorithms for testing if a polynomial is cyclotomic.
