Problem about squares

[Peter Pein]

Well, then take 2^(n+1) for even n and 2^(n+1)+1 for odd n if this is not a
square, else for odd n try 4^k*2^(n+1)+1 for k from 1 to somewhere until the
result is not a square.
.
Joshua Zucker schrieb:
> I think you two are talking about different questions: Peter shows
> it's not true that there exists such a number that works for all n but
> David wants to know if, for each n, there exists a (not necessarily
> the same) number.
> 
> --Joshua Zuker
> 
> On Jan 11, 2008 7:51 AM, Peter Pein <petsie at dordos.net> wrote:
>> David W. Wilson schrieb:
>>> Does there exist a non-square number which is square (mod 2^n) for every
>>> n >= 0?
>>>
>> Call this non-square a0 and have a look at n0=ceil(log(a0)/log(2)). for every
>> n>=n0: a0 == a0 mod 2^n and a0 has to be square and non-square simultaneously.
>> cantradiction.
>>
>> P²
>>
> 
>