Question 1208953
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Suppose a &nbsp;<U>is a natural number</U>&nbsp; >= 2 and n is a natural number larger than 1.
How can I prove that if &nbsp;n &nbsp;is odd, &nbsp;then &nbsp;{{{a^n+1}}} &nbsp;is not a prime?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I edited your post - see the underlined fragment - to make it precisely correct.



<pre>
If n is an odd positive integer number greater than 1, then there is a decomposition


    {{{a^n + 1}}} = {{{(a+1)*(a^(n-1) - a^(n-2) + ellipsis - a + 1)}}}


(the signs at degrees of "a" alternate " + " and " - ").


It is a well known formula. To prove it, it is enough to open parentheses.


This formula shows and tells that every integer number of the form  {{{a^n+1}}} 
with integer positive  " a "  and natural odd  n > 1  is a composite number.


It is what you want to prove.
</pre>

Solved.



For &nbsp;n=3, &nbsp;the decomposition  &nbsp;{{{a^3+1}}} = {{{(a+1)*(a^2 - a + 1)}}} &nbsp;is studied explicitly 
in standard school &nbsp;Math curriculum as the sum of cubes decomposition.



For other odd natural integer &nbsp;" n " &nbsp;greater than &nbsp;3, &nbsp;it can be 
easily obtained from the formula of the sum of a geometric progression

    1, &nbsp;-a, &nbsp;a^2, &nbsp;-a^3, &nbsp;a^4, &nbsp;-a^5, . . . , &nbsp;a^(n-1)


with alternate signs, &nbsp;which corresponds to the case of common ratio &nbsp;" -a ".