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Arithmetics in Z

  1. Sep 24, 2011 #1
    Hello all,

    I have a problem to define a set of natural numbers that meet the following equation:

    m2+1[itex]\equiv[/itex]0[5]

    I have found that a set of this equation is : {2,3,7,8}+k*10, k[itex]\in[/itex]N
    i.e: k= {0,1,2,3,...........,n}
    example: (2+(0*10)2)+1=5
    (3+(1*10)2)+1=132+1=170

    Result: m=[2+k*10; 3+k*10; 7+k*10; 8+k*10]

    How can I describe this result logically in mathematics ?

    Thanks.
     
  2. jcsd
  3. Sep 24, 2011 #2
    Hello,
    I think you more or less did: m is congruent to either 2, 3, 7 or 8 modulo 10.

    Notice that this is equivalent to saying that m is congruent to either 2 or 3 modulo 5, since [itex]2 \equiv 7 \pmod 5[/itex] and [itex]3 \equiv 8 \pmod 5[/itex].

    You wanted a solution of the equation [itex]m^2 \equiv -1 \equiv 4 \pmod 5[/itex]; you only need to square each of 0,1,2,3,4 modulo 5, and see which are congruent to 4.

    Hope this helps!
     
  4. Sep 24, 2011 #3
    Look at sequence A047221 in OEIS ('Numbers that are congruent to {2, 3} mod 5')
     
  5. Sep 29, 2011 #4
    Hello,
    thank you for your reply.
    I have found a solution :
    m2+1 [itex]\equiv[/itex] 0[5]
    m2 [itex]\equiv[/itex] -1[5]
    m2 [itex]\equiv[/itex] 4[5]
    m [itex]\equiv[/itex] [itex]\pm[/itex]2[5]
    m [itex]\equiv[/itex] 2,3[5]

    therefore : m= {2+5n ; 3+5n} for any integer n[itex]\geq[/itex]0
     
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