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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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}$$x^2 equiv -2,2 pmod {122}$Method to find solution for $a^x equiv mod n$
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Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?
Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?
Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.
The first congruence has no solution below $10^9$
number-theory elementary-number-theory modular-arithmetic
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add a comment |
$begingroup$
Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?
Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?
Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.
The first congruence has no solution below $10^9$
number-theory elementary-number-theory modular-arithmetic
$endgroup$
$begingroup$
$$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
$endgroup$
– Peter
Mar 27 at 18:48
add a comment |
$begingroup$
Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?
Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?
Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.
The first congruence has no solution below $10^9$
number-theory elementary-number-theory modular-arithmetic
$endgroup$
Related to this question : Can I find all solutions of $2^{n-1}equiv kmod n$?
Does one of the congruences $$2^{n-1}equiv 2^{16}+1mod n$$ and $$2^{n-1}equiv 2^{26}+1mod n$$ have an integer solution $n>1$ ?
Enzo Creti checked the second congruences upto $ 127cdot 10^9 $. No solution was found.
The first congruence has no solution below $10^9$
number-theory elementary-number-theory modular-arithmetic
number-theory elementary-number-theory modular-arithmetic
edited Mar 22 at 13:01
Peter
asked Mar 22 at 12:47
PeterPeter
49.1k1240138
49.1k1240138
$begingroup$
$$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
$endgroup$
– Peter
Mar 27 at 18:48
add a comment |
$begingroup$
$$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$$
$endgroup$
– Peter
Mar 27 at 18:48
add a comment |
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$begingroup$
$$n = 4 428 169 422 323$$ solves $$2^{n-1}equiv 2^{16}+1mod n$$
$endgroup$
– Peter
Mar 27 at 18:48