In base 10, 1364 is not a perfect square. What is the smallest base for which 1364 IS a perfect square?

Question

Hint: The base must be at least 7 for 1364 to make sense in that base.

I want the number whose perfect square is represented by the digits 1-3-6-4 in the base you found.

And don't forget to tell me the base :)

Answer 1

Let the base be equal to b. Then

b^3+3b^2+6b+1 = s^2 = (b+1)(b^2+2b+4)=(b+1)((b+1)^2+3)

forcing 3 divides (b+1) lest (b+1)^2 +3 be a square and b=0.

This gives s^2 = (9)((b+1)/3)((b+1)^2/3 + 1) so setting

(b+1)/3 = p; s^2 = 9p (3p^2 + 1) giving squares

p=u^2 and 3p^2 + 1 = v^2 so that 3u^4 + 1 = v^2 which

has smallest solutions at u=1 , u=2 for which b= 3u^2-1

at u=1,2 giving b = 2,11 but b>6 finally gives b=11.

Answer 2

It's base 11. The value of the number will be a perfect square in any base. That's because it's square root is an integer in any base.

1364 = 4 + 6n + 3n^2 + n^3
n = 11 gives 42^2 in base 10

So the square root of the number 1364_11 is 39_11

Answer 3

You may have addled notation in base 2, but if you do the calculations, 1364 in base 2 gives you 36, which is a perfect square. (You get 4 plus 6*2 plus 3*4, plus 1*8, which totals 36.)

It's also a perfect square in base 11.

For what it's worth, it's not a perfect square in base 12 through base 100. I didn't check beyond 100.

Answer 4

you want the number whose perfect square is represented by whatever value is represented by the digits 1-3-6-4 ?

Or do you want to convert the base ten number 1364 into a different base before figuring what its perfect square would be

Answer 5

Base 11
4 + 6*11 + 3*11^2 + 1*11^3 = 1764 (all in base 10)
1764 = 42^2

Answer 6

In base b >=7, 1364_b = b^3 + 3b^2 + 6b + 4. Well, I just checked and concuded the smallest base is 11, because 1364_b = 1764 = 42^2.

Answer 7

I think base 11 is the only base where it's a perfect square -- the b^3 dominates, and isn't close to an analytical square. No other bases below 100,000,000 work.

Answer 8

why?