2005 base 10 is not a square. Neither is 2005 base 7 a square (equal 2*7^3+5=691). Is there any base b such that 2005 base b is a square?

x^2 = 2 * b^3 + 5 

If the base b is odd, then we can write b = 2m+1.

x^2 = 2 * (2m+1)^3 + 5

x^2 = 2 * (8m^3 + 12m^2 + 6m + 1) + 5

x^2 = 16m^3 + 24m^2 + 12m + 7

x^2 = 4m(4m^2 + 3) + 6(4m^2 + 3) - 11

x^2 = (4m + 6)(4m^2 + 3) - 11

x^2 + 11 = (4m + 6)(4m^2 + 3)

4m + 6 is congruent to 2 mod 4.

4m^2 + 3 is congruent to 3 mod 4.

(4m + 6)(4m^2 + 3) is congruent to 2x3 mod 4 or 2 mod 4.

x^2 + 11 is congruent to x^2 + 3 mod 4.

x^2 + 3 (must be congruent to) 2 mod 4.

x^2  (must be congruent to) -1 mod 4.

x^2 (must be congruent to) 3 mod 4.

A square is congruent (mod 4) to 0 or 1, never congruent to 3.

So, if there exists a base, it's not odd.

Am I wrong ?

I'm working yet.  

 

Posted by pcbouhid on 2005-05-30 20:10:36