Is there a solution for $2^{n-1}equiv 2^{16}+1mod n$ or $2^{n-1}equiv 2^{26}+1mod n$? The 2019...

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Is there a solution for $2^{n-1}equiv 2^{16}+1mod n$ or $2^{n-1}equiv 2^{26}+1mod n$?



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Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Can I find all solutions of $2^{n-1}equiv kmod n$?$6^x equiv 11 mod{17}$Find minimal $xinBbb N$ that solves the linear congruenceHow to simplify the modular congruence: $4kequiv 4(text{mod } 16)$Show that there exists an integer $x$ such that $xequiv 23 mod 1000$ and $xequiv 45 mod 6789$Compute $50! mod 2014$Solve $x^2equiv a$ mod $prod P_i^{e_i}$Solve $y^4 equiv 5 bmod{21} $Solutions of $x^2-6x-13 equiv 0 pmod{127}$$x^2 equiv -2,2 pmod {122}$Method to find solution for $a^x equiv mod n$












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$begingroup$


Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?




Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?




Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.



The first congruence has no solution below $10^9$










share|cite|improve this question











$endgroup$












  • $begingroup$
    $$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
    $endgroup$
    – Peter
    Mar 27 at 18:48
















1












$begingroup$


Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?




Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?




Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.



The first congruence has no solution below $10^9$










share|cite|improve this question











$endgroup$












  • $begingroup$
    $$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
    $endgroup$
    – Peter
    Mar 27 at 18:48














1












1








1


1



$begingroup$


Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?




Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?




Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.



The first congruence has no solution below $10^9$










share|cite|improve this question











$endgroup$




Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?




Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?




Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.



The first congruence has no solution below $10^9$







number-theory elementary-number-theory modular-arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 13:01







Peter

















asked Mar 22 at 12:47









PeterPeter

49.1k1240138




49.1k1240138












  • $begingroup$
    $$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
    $endgroup$
    – Peter
    Mar 27 at 18:48


















  • $begingroup$
    $$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
    $endgroup$
    – Peter
    Mar 27 at 18:48
















$begingroup$
$$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
$endgroup$
– Peter
Mar 27 at 18:48




$begingroup$
$$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
$endgroup$
– Peter
Mar 27 at 18:48










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