Playing with bases

[BMAD]

Suppose n is an integer in base 10. Note that n² ends with 0 if and only if n ends with 0. Now consider integers represented in base b, where 5 <= b <= 9. Determine for which base b (if any) the following statement is true:

For any integer n, n² ends with 0 if and only if n ends with 0.

[vistaptb]

Basically, the question is, does a number have to be a multiple of (insert base here) to have a square that is a multiple of (insert base here)? I guess that statement would apply to

5, 6, and 7 (4 squared is 20 in base 8, and 3 squared is 10 in base 9).

[phil1882]

i'm pretty sure that statement is true for any base.

base 5: 10^2 = 100. (5^2 = 25)

base 6: 10^2 = 100. (6^2 = 36)

etc.

a number will end in zero in a particular base if and only if it's evenly divisable by the base value. and squaring a number doesn't change its divisablty with respect to the base.

[ParaLogic]

i'm pretty sure that statement is true for any base.

base 5: 10^2 = 100. (5^2 = 25)

base 6: 10^2 = 100. (6^2 = 36)

etc.

a number will end in zero in a particular base if and only if it's evenly divisable by the base value. and squaring a number doesn't change its divisablty with respect to the base.

It's true that if you start with a number that ends with 0, its square will also end with 0. But the OP asks for a case of

if and only if, meaning that if you can find a number that ends with 0, while its (integer) square root does not end with 0, then the statement is not true for that base. In his post, vistaptb gave two such occurences.

[phil1882]

okay i see. i was reading "only if" more literally rather than its actual inverse statement.

[witzar]

The statement is false for every base that is divisible by a square of some prime and true otherwise.

So it it is false for 8, 9 and true for 5, 6, 7 (and 10).

[kingofpain]

This is only true for numbers whose prime factorization leads to a list of unique numbers...