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Let $$f(n):=n^{n^2}+(n+1)^{(n+1)^2}$$ for a positive integer $n$. For clarification $n^{n^2}$ means $n^{(n^2)}$ , analogue for the other summand.

Can we find a concrete factor of $\ f(n)\ $ (like algebraic or aurifeuillan factors) ?

Motivation : I still could not completely factor $f(10)$ (there is a $114$ digit composite left) and I know no factor of the $309$-digit composite number $f(15)$.

See also here : http://factordb.com/index.php?query=n%5E%28n%5E2%29%2B%28n%2B1%29%5E%28%28n%2B1%29%5E2%29

I wonder whether we can use any trick to find the factors easier than by simply using the standard techniques , for example , can we make us of the structure of those numbers if we use the quadratic sieve ?

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    $\begingroup$ The exponents are coprime, and are not independent of the bases, so that's worrying. $\endgroup$
    – anon
    Commented Aug 12, 2019 at 10:54
  • $\begingroup$ It need not be clarified, it's convention @Peter. Parentheses are only needed, when you want to force a different order, or to group things. $\endgroup$
    – user645636
    Commented Aug 12, 2019 at 12:45
  • $\begingroup$ It's congruent to $n+1$ or $n+2$ mod numbers with Euler phi of 4. Depends on if n is odd, or even respectively. $\endgroup$
    – user645636
    Commented Aug 12, 2019 at 17:43
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    $\begingroup$ If there were nice factors, then probably $f(2)$ wouldn't be prime. $\endgroup$
    – Ivan Neretin
    Commented Aug 12, 2019 at 18:14
  • $\begingroup$ @Ivan f(1) is also prime. However The Mersenne sequence has a two prime start as well. $\endgroup$
    – user645636
    Commented Aug 12, 2019 at 18:59

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