In which postitive number base do the numerals 13, 38, and 41 form a Pythagorean Triplet?

Question

a2 + b2 = c2
2 - squared

Answer 1

Let the base be k.

13 base k = k + 3.
38 base k = 3k + 8.
41 base k = 4k + 1.

Now use the Pythagorean Theorem.

(k+3)^2 + (3k + 8)^2 = (4k + 1)^2

k^2 + 6k + 9 + 9k^2 + 48k + 64 = 16k^2 + 8k + 1
10k^2 + 54k + 73 = 16k^2 + 8k + 1
0 = 6k^2 - 46k - 72
0 = 3k^2 - 23k - 36
0 = (3k + 4)(k - 9)

k = -4/3, or k = 9.

The base must be base 9.

Let's check:

13 base 9 = 12 base 10.
38 base 9 = 35 base 10
41 base 9 = 37 base 10


12^2 = 144, 35^2 = 1225, 37^2 = 1369
144 + 1225 = 1369.

It works!

Answer 2

Well, since one of the numbers is 38, the base has to be at least 9. (For any base n, the maximum digit allowed is n-1.)

It turns out that 9 works.

13 base 9 = 1x9 + 3 = 12
38 base 9 = 3x9 + 8 = 35
41 base 9 = 4x9 + 1 = 37

And 12^2 + 35^2 = 37^2.

By the way, props to hokiejthweatt. His way is the best way to do it in general. My way was just to try the first base that MIGHT work...and it worked, but it wouldn't always.

Answer 3

That certainly is an unusual question. Reminds me of an exchange in 'Alice in Wonderland' that could only be understood in the context of number bases.

Answer 4

Dang, hokie, nice solution!