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Small solutions to modular arithmetic linear congruence


How does NaCl Poly1305 implementation do modular multiplication?Understanding elliptic curve encryptionFermats Little Theorem, primitive rootPolynomial ModulusFactoring large $N$ given oracle to find square roots modulo $N$How is it possible that $g^q equiv 1 pmod p$ for a generator g?Given a prime exponent e and a prime number n, find b, where b^e = 1 mod nIncorrect solution for Discrete Log Problem when using the Index Calculus algorithmDiscrete logarithm weak groupSolving the discrete logarithm problem for a weak group













1












$begingroup$


Let $p$ be a prime number with $N$ bits, let $a,b,c$ be constants. The problem is to find solutions to the equivalent $a x + b y equiv c pmod p$ with both having at most $N/2$ bits.



What algorithmic approaches can solve this problem? Does it have any known hardness reduction?






modular-arithmetic







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
    $endgroup$
    – Marc Ilunga
    9 hours ago
















1












$begingroup$


Let $p$ be a prime number with $N$ bits, let $a,b,c$ be constants. The problem is to find solutions to the equivalent $a x + b y equiv c pmod p$ with both having at most $N/2$ bits.



What algorithmic approaches can solve this problem? Does it have any known hardness reduction?






modular-arithmetic







share|improve this question









$endgroup$








  • 1




    $begingroup$
    Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
    $endgroup$
    – Marc Ilunga
    9 hours ago














1












1








1


1



$begingroup$


Let $p$ be a prime number with $N$ bits, let $a,b,c$ be constants. The problem is to find solutions to the equivalent $a x + b y equiv c pmod p$ with both having at most $N/2$ bits.



What algorithmic approaches can solve this problem? Does it have any known hardness reduction?






modular-arithmetic







share|improve this question









$endgroup$




Let $p$ be a prime number with $N$ bits, let $a,b,c$ be constants. The problem is to find solutions to the equivalent $a x + b y equiv c pmod p$ with both having at most $N/2$ bits.



What algorithmic approaches can solve this problem? Does it have any known hardness reduction?






modular-arithmetic




modular-arithmetic






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 11 hours ago









rain1rain1

1106




1106








  • 1




    $begingroup$
    Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
    $endgroup$
    – Marc Ilunga
    9 hours ago














  • 1




    $begingroup$
    Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
    $endgroup$
    – Marc Ilunga
    9 hours ago








1




1




$begingroup$
Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
$endgroup$
– Marc Ilunga
9 hours ago




$begingroup$
Coppersmith's methods are generally used to solve this types of problems. I can't write a proper answer now but have a look at this paper: cits.rub.de/imperia/md/content/may/paper/jochemszmay.pdf
$endgroup$
– Marc Ilunga
9 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

You can use lattice reduction to solve this problem.



Pick a large constant $Sinmathbb Z$ and consider the lattice spanned by the rows of the following matrix:
$$
L = begin{pmatrix}
S a & -1 & 0 & 0 \
S b & 0 & -1 & 0 \
S c & 0 & 0 & S \
S p & 0 & 0 & 0 \
end{pmatrix}
$$



Now the crucial thing to notice is that some pair $(x,y)inmathbb Z^2$ is a solution to your modular equation if and only if $(0,x,y,S)$ is a vector in this lattice.



Moreover, some vector of the form $vec v=(Sz,x,y,pm S)$ must be part of a short basis, since $begin{pmatrix}S c & 0 & 0 & Send{pmatrix}$ is the only row of $L$ that is non-zero in the last column. Due to the large scaling factor $S$ in the first column, the vector $vec v$ will in fact satisfy $z=0$, and therefore you can find a short solution by computing a reduced basis of $L$.



Here's a sage transcript that demonstrates this:





sage: p = next_prime(2**32)

sage: N = 1+floor(log(p,2)) # bit length

sage: S = 10**N

sage: a, b, c = randrange(p), randrange(p), randrange(p)

sage: a, b, c

(2206104035, 3690588304, 373686466)

sage: L = matrix(ZZ, [[S*a,-1,0,0], [S*b,0,-1,0], [S*c,0,0,S], [S*p,0,0,0]])

sage: L

[22061040350000000000                   -1                    0                    0]

[36905883040000000000                    0                   -1                    0]

[ 3736864660000000000                    0                    0          10000000000]

[42949673110000000000                    0                    0                    0]

sage: L.LLL()

[           0        49124        -7835            0]

[           0       -31049       -82479            0]

[-10000000000         2330       -37438            0]

[           0         4276       -42601  10000000000]

sage: (4276*a -42601*b) % p == c

True





share|improve this answer









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    active

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    3












    $begingroup$

    You can use lattice reduction to solve this problem.



    Pick a large constant $Sinmathbb Z$ and consider the lattice spanned by the rows of the following matrix:
    $$
    L = begin{pmatrix}
    S a & -1 & 0 & 0 \
    S b & 0 & -1 & 0 \
    S c & 0 & 0 & S \
    S p & 0 & 0 & 0 \
    end{pmatrix}
    $$



    Now the crucial thing to notice is that some pair $(x,y)inmathbb Z^2$ is a solution to your modular equation if and only if $(0,x,y,S)$ is a vector in this lattice.



    Moreover, some vector of the form $vec v=(Sz,x,y,pm S)$ must be part of a short basis, since $begin{pmatrix}S c & 0 & 0 & Send{pmatrix}$ is the only row of $L$ that is non-zero in the last column. Due to the large scaling factor $S$ in the first column, the vector $vec v$ will in fact satisfy $z=0$, and therefore you can find a short solution by computing a reduced basis of $L$.



    Here's a sage transcript that demonstrates this:





    sage: p = next_prime(2**32)
    
sage: N = 1+floor(log(p,2)) # bit length
    
sage: S = 10**N
    
sage: a, b, c = randrange(p), randrange(p), randrange(p)
    
sage: a, b, c
    
(2206104035, 3690588304, 373686466)
    
sage: L = matrix(ZZ, [[S*a,-1,0,0], [S*b,0,-1,0], [S*c,0,0,S], [S*p,0,0,0]])
    
sage: L
    
[22061040350000000000                   -1                    0                    0]
    
[36905883040000000000                    0                   -1                    0]
    
[ 3736864660000000000                    0                    0          10000000000]
    
[42949673110000000000                    0                    0                    0]
    
sage: L.LLL()
    
[           0        49124        -7835            0]
    
[           0       -31049       -82479            0]
    
[-10000000000         2330       -37438            0]
    
[           0         4276       -42601  10000000000]
    
sage: (4276*a -42601*b) % p == c
    
True





    share|improve this answer









    $endgroup$


















      3












      $begingroup$

      You can use lattice reduction to solve this problem.



      Pick a large constant $Sinmathbb Z$ and consider the lattice spanned by the rows of the following matrix:
      $$
      L = begin{pmatrix}
      S a & -1 & 0 & 0 \
      S b & 0 & -1 & 0 \
      S c & 0 & 0 & S \
      S p & 0 & 0 & 0 \
      end{pmatrix}
      $$



      Now the crucial thing to notice is that some pair $(x,y)inmathbb Z^2$ is a solution to your modular equation if and only if $(0,x,y,S)$ is a vector in this lattice.



      Moreover, some vector of the form $vec v=(Sz,x,y,pm S)$ must be part of a short basis, since $begin{pmatrix}S c & 0 & 0 & Send{pmatrix}$ is the only row of $L$ that is non-zero in the last column. Due to the large scaling factor $S$ in the first column, the vector $vec v$ will in fact satisfy $z=0$, and therefore you can find a short solution by computing a reduced basis of $L$.



      Here's a sage transcript that demonstrates this:





      sage: p = next_prime(2**32)
      
sage: N = 1+floor(log(p,2)) # bit length
      
sage: S = 10**N
      
sage: a, b, c = randrange(p), randrange(p), randrange(p)
      
sage: a, b, c
      
(2206104035, 3690588304, 373686466)
      
sage: L = matrix(ZZ, [[S*a,-1,0,0], [S*b,0,-1,0], [S*c,0,0,S], [S*p,0,0,0]])
      
sage: L
      
[22061040350000000000                   -1                    0                    0]
      
[36905883040000000000                    0                   -1                    0]
      
[ 3736864660000000000                    0                    0          10000000000]
      
[42949673110000000000                    0                    0                    0]
      
sage: L.LLL()
      
[           0        49124        -7835            0]
      
[           0       -31049       -82479            0]
      
[-10000000000         2330       -37438            0]
      
[           0         4276       -42601  10000000000]
      
sage: (4276*a -42601*b) % p == c
      
True





      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        You can use lattice reduction to solve this problem.



        Pick a large constant $Sinmathbb Z$ and consider the lattice spanned by the rows of the following matrix:
        $$
        L = begin{pmatrix}
        S a & -1 & 0 & 0 \
        S b & 0 & -1 & 0 \
        S c & 0 & 0 & S \
        S p & 0 & 0 & 0 \
        end{pmatrix}
        $$



        Now the crucial thing to notice is that some pair $(x,y)inmathbb Z^2$ is a solution to your modular equation if and only if $(0,x,y,S)$ is a vector in this lattice.



        Moreover, some vector of the form $vec v=(Sz,x,y,pm S)$ must be part of a short basis, since $begin{pmatrix}S c & 0 & 0 & Send{pmatrix}$ is the only row of $L$ that is non-zero in the last column. Due to the large scaling factor $S$ in the first column, the vector $vec v$ will in fact satisfy $z=0$, and therefore you can find a short solution by computing a reduced basis of $L$.



        Here's a sage transcript that demonstrates this:





        sage: p = next_prime(2**32)
        
sage: N = 1+floor(log(p,2)) # bit length
        
sage: S = 10**N
        
sage: a, b, c = randrange(p), randrange(p), randrange(p)
        
sage: a, b, c
        
(2206104035, 3690588304, 373686466)
        
sage: L = matrix(ZZ, [[S*a,-1,0,0], [S*b,0,-1,0], [S*c,0,0,S], [S*p,0,0,0]])
        
sage: L
        
[22061040350000000000                   -1                    0                    0]
        
[36905883040000000000                    0                   -1                    0]
        
[ 3736864660000000000                    0                    0          10000000000]
        
[42949673110000000000                    0                    0                    0]
        
sage: L.LLL()
        
[           0        49124        -7835            0]
        
[           0       -31049       -82479            0]
        
[-10000000000         2330       -37438            0]
        
[           0         4276       -42601  10000000000]
        
sage: (4276*a -42601*b) % p == c
        
True





        share|improve this answer









        $endgroup$



        You can use lattice reduction to solve this problem.



        Pick a large constant $Sinmathbb Z$ and consider the lattice spanned by the rows of the following matrix:
        $$
        L = begin{pmatrix}
        S a & -1 & 0 & 0 \
        S b & 0 & -1 & 0 \
        S c & 0 & 0 & S \
        S p & 0 & 0 & 0 \
        end{pmatrix}
        $$



        Now the crucial thing to notice is that some pair $(x,y)inmathbb Z^2$ is a solution to your modular equation if and only if $(0,x,y,S)$ is a vector in this lattice.



        Moreover, some vector of the form $vec v=(Sz,x,y,pm S)$ must be part of a short basis, since $begin{pmatrix}S c & 0 & 0 & Send{pmatrix}$ is the only row of $L$ that is non-zero in the last column. Due to the large scaling factor $S$ in the first column, the vector $vec v$ will in fact satisfy $z=0$, and therefore you can find a short solution by computing a reduced basis of $L$.



        Here's a sage transcript that demonstrates this:





        sage: p = next_prime(2**32)
        
sage: N = 1+floor(log(p,2)) # bit length
        
sage: S = 10**N
        
sage: a, b, c = randrange(p), randrange(p), randrange(p)
        
sage: a, b, c
        
(2206104035, 3690588304, 373686466)
        
sage: L = matrix(ZZ, [[S*a,-1,0,0], [S*b,0,-1,0], [S*c,0,0,S], [S*p,0,0,0]])
        
sage: L
        
[22061040350000000000                   -1                    0                    0]
        
[36905883040000000000                    0                   -1                    0]
        
[ 3736864660000000000                    0                    0          10000000000]
        
[42949673110000000000                    0                    0                    0]
        
sage: L.LLL()
        
[           0        49124        -7835            0]
        
[           0       -31049       -82479            0]
        
[-10000000000         2330       -37438            0]
        
[           0         4276       -42601  10000000000]
        
sage: (4276*a -42601*b) % p == c
        
True






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 8 hours ago









        yyyyyyyyyyyyyy

        9,65933452




        9,65933452






























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