Solution for Arithmetics in Z: m2+1\equiv0[5]

[naoufelabs]

Hello all,

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

m²+1\equiv0[5]

I have found that a set of this equation is : {2,3,7,8}+k*10, k\inN
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.

 

[dodo]

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 2 \equiv 7 \pmod 5 and 3 \equiv 8 \pmod 5.

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

Hope this helps!

 

[RamaWolf]

Look at sequence A047221 in OEIS ('Numbers that are congruent to {2, 3} mod 5')

 

[naoufelabs]

Hello,
thank you for your reply.
I have found a solution :
m²+1 \equiv 0[5]
m² \equiv -1[5]
m² \equiv 4[5]
m \equiv \pm2[5]
m \equiv 2,3[5]

therefore : m= {2+5n ; 3+5n} for any integer n\geq0