Perfect square in 2 bases

1. Aug 9, 2009

srijithju

Does anyone know if its possible to check whether a number is a perfect square in 2 different bases ? ( I dont mean the representation of a number in 2 different bases - coz that would be a ridiculous question , what I mean is consider a no say xyz - is it possible to prove / disprove that it is a perfect square when considered in 2 bases .. i.e. lets say bases m and n ..... then the numbers would be x * m^2 + y * m + z in base m and x* n^2 + y*n + z in base n) .

2. Aug 9, 2009

CRGreathouse

If a sequence of digits is a square in some base it should be possible to find that base by exhaustive testing. If not, I'm not sure how to prove it -- unless perhaps it's close to a perfect power, like 101_b = b^2 + 1.

3. Aug 10, 2009

srijithju

What I am looking for is basically an algorithm ... Suppose I find one such base by exhaustive testing , now is it possible to find another base , or do I have to do exhaustive testing to find the second base too .

Or is there some way to rule out a certain sequence of digits from being a perfect square in all possible bases , or maybe is it always possible to find a base in which a sequence of digits will be a perfect square, given that the sequence of digits satisfy some relation .

All I am able to see is the following :

1. that if the sequence has only 1 non zero digit :
Here there are 2 scenarios -
1st scenario : in case that the sequence has an even no. of digits ( when we do not consider the leading zeroes before the non zero digit of course) then u can always find a base in which the number is a perfect square ..

eg. p000 , where p is any digit ( or even a letter representing a single digit in bases higher than 10)

then in a base q the no. would be :
= p * q^3
= p*q * q^2

Now if q = (p*any perfect square)
, then we can clearly see the number is a perfect square in base q , i.e we have an infinite no. of bases in which the number is a perfect square

2nd scenario : the sequence has odd no. of digits - In this case the only way for the num to be a square in any arbitrary base is if the non zero digit itself is a perfect square - this can be seen similar to above.

2. If the digits are in a Geometric progression :

If the no. is (k digits):
r(k-1) r(k) ... r0

If let us say the common ratio is r , then in a base q , the num would be :

r0 * { (r*q)^k - 1 } / {r - 1}

Now we require the num to be a square , so the above num should be a square,
If now somehow we find integer solns of q by equating the above to a square , then we can find the bases ,
If we choose a square no x^2 ,
then we have :
r0 * { (r*q)^k - 1 } / {r - 1} = x^2

well I guess this just leads us to an even bigger problem to solve ....

nyways can anyone find some relation , that may help in determining the solution ?

4. Aug 10, 2009

ramsey2879

400 is a perfect square in any base n
961 is also a perfect square in any base n

Last edited: Aug 10, 2009
5. Aug 10, 2009

Mensanator

400 does not exist in any base n < 5.

961 does not exist in any base n < 10.

6. Aug 11, 2009

ramsey2879

You were just a hair off. p000 is a perfect square in any base $$p^{2n+1}$$ n > 0