In which base number system is $23^2=562$?

Question

This is an exercise from a math chapter called number systems where I learn how to convert decimal numbers to other number bases like binary and hexadecimal.

Here I need to discover which base number system is $23^2 = 562$. I know that the resulting base number system is $7$. Since $23$ septenary = $17$ decimal, and $17^2$ decimal = $289$, I can confirm that $289$ decimal = $562$ septenary.

The problem is that I must discover $7$ using equations and not with trial-and-error. Thank you!

Answer 1

So that this question can be marked as answered:

In base $b$, $23^2 = (2b + 3)^2 = 4b^2 + 12b + 9$. Equate this to $5b^2 + 6b + 2$ to get $b^2 - 6b - 7 = 0$, which can now be solved to get $b = 7$.

Answer 2

Well you can try guessing but...

$23^2 = (2\times b + 3)^2 = 4b^2 + 12b + 9$

And $562 = 5b^2 + 6b + 2$.

So $4b^2 + 12b + 9 = 5b^2 + 6b + 2$.

We have several options now.

Maybe straightforward is simply solve the quadratic.

$b^2-6b-7 = 0$ so $b =\frac {6\pm \sqrt{36+4\cdot 7}}2 = -1, 7$ so $b=7$.

....

But that doesn't really teach us to think like number theorists.

We do knot that if $23_b = X$ then $X \equiv 3\pmod b$ and $X^2 \equiv 9\pmod b$.

But $X^2 = 562_b$ so $X^2\equiv 2\pmod b$. So $2 \equiv 9\pmod b$. So $2$ and $9$ differe by a multiple of $b$. That is $b| 9-2=7$ so $b|7$ but $7$ is prime so $b = 7$ is the only option.

We can verify $b=7$ works for the rest....

.....

Alternatively to get a look at "what is going on" we see that

$562_b = 5b^2 + 6b + 2$. But $23_b^2 = 4b^2 + 12b+9$

So we have $5b^2 + 6b + 2 = 4b^2 + 12b +9$.

$5b^2 + 6b + 2 = 562_b$ is the "correct" written which means that $4b^2 + 12b+9$ is "out of balance and we need to carry or borrow somehow.

So we need to turn $4b^2 + 12b + 9$ into $5b^2 + 12b + 9$ by borrowing or carrying.

$4b^2 + 12b + 9=$
$5b^2 - b^2 + 12b + 9 =$
$5b^2 - b\cdot b + 12b + 9 = $
$5b^2 + (12-b)b + 9$.

Now we need to turn $(12-b)b$ to $6b$ by borrowing and carrying.

$5b^2 + (12 - b)b + 9 = $
$5b^2 + (6 + 6-b)b + 9 = $
$5b^2 + 6b + [9 + (6-b)b]$.

This leaves us with $9 + (6-b)b = 2$ and we can solve for $b$ with the quadratic equation.

But it probably have been easier if back when we had

$5b^2 + (12-b)b + 9$ if we had realized we have to turn the $9$ into a $2$.

$5b^2 + (12-b)b + 9=$
$5b^2 + (12-b)b + (7+2) =$
$5b^2 + (12 - b+\frac 7b)b + 2$

So $12 - b + \frac 7b = 6$.

Now $\frac 7b$ must be an integer so we must have $b|7$ so we must have $b=1$ or $7$ and if we have $b = 7$ we get $12 - 7 +\frac 7b = 6$

....

or we could have gone the other way from the start.

$4b^2 + 12b + 9 = 5b^2 + 12b + 9$ so

$4b^2 + 12b +9 = $
$4b^2 + 12b + (7+2) =$
$4b^2 + (12 + \frac 7b)b + 2=$
$4b^2 + (6+\frac 7b + 6)b + 2 =$
$(4+\frac 6b + \frac 7{b^2})b^2 + 6b +2=$
$5b^2 + 6b + 2$ if and only if

$\frac 6b + \frac 7{b^2} = 1$.

So $6b + 7 = b^2$. So solve for $b$ and we $b = 7$ or $b=-1$.

....

However you do it though, the key is we have $23^2 = 4b^2 + 12b +9= 5b^2 + 6b + 2$ so we must have $9\equiv 2\pmod b$ and that limits what $b$ can be (in this case $b$ must be $7$)