We know that $23$ is a prime number nonetheless, I'm asking to find 4 numbers $a,b,c,d > 0$ such that $23$ factors.

$$ 23 = A \times B \times C \times D \text{ with } A,B,C,D = a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6}$$

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  • all A, B, C, and D have the same format a+ ... with different values for a, b, c, d? – Marius Apr 19 at 14:50
  • 1
    the format should be $A = a + b\sqrt{2} + c\sqrt{3} + d \sqrt{6}$ but perhaps not the same $a,b,c,d$ for each one. – john mangual Apr 19 at 14:53
  • Do a,b,c,d have to be integers? – Ankoganit Apr 19 at 14:54
  • @Ankoganit. Most probably no. Making a=b=c=d=1 will get you A*B*C*D way over 23. – Marius Apr 19 at 15:00
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    You are talking about prime numbers and number theory, so it seems like you want $a,b,c,d$ to be integers, but the constraint $a,b,c,d>0$ would make this impossible. Perhaps you meant $a,b,c,d\neq 0$? – Mike Earnest Apr 19 at 17:27

Assuming a,b,c,d are meant to be integers (which makes the problem much more interesting) then the answer is


This reduces to a difference of two squares


If $a,b,c,d\neq 0$ then one of many solutions is:


To see this and figure it out, you need a couple of steps.

Firstly, you go for a difference of two squares $\pm23=(k+j\sqrt{n})(k-j\sqrt{n})$. We follow a vague intuition that $n=6$ based on the observation that $\sqrt{2}\sqrt{3}=\sqrt{6}$. You will therefore need $\pm23=k^2-6j^2$.

By inspection, $23=24-1=(2\sqrt{6}-1)(2\sqrt{6}+1)$, so this seems like a reasonable place to start.

Next, consider $(a_1+b_1\sqrt{2}+c_1\sqrt{3}+d_1\sqrt{6})(a_2+b_2\sqrt{2}+c_2\sqrt{3}+d_2\sqrt{6})$.

We want this to equal $2\sqrt{6}\pm1$. If we multiply it out, we get

\begin{eqnarray} a_1a_2&+&2b_1b_2+3c_1c_2+6d_1d_2\\ &+&\sqrt{2}(a_1b_2+b_1a_2)\\ &+&\sqrt{3}(a_1c_2+c_1a_2)\\ &+&\sqrt{6}(a_1d_2+d_1a_2+b_1c_2+c_1b_2)\\ &+&\sqrt{12}(b_1d_2+d_1b_2)\\ &+&\sqrt{18}(a_1b_2+b_1a_2) \end{eqnarray}

It is clear that we need to get rid of the cross-products of $a\times b$, $a\times c$, $d\times b$, and $d\times c$ (where I mean by this the co-efficients of $\sqrt{2},\sqrt{3},\sqrt{12},\sqrt{18}$). This is easily done by having $a_1=\pm a_2$, $b_1=\pm b_2$, $c_1=\pm c_2$, and $d_1=\pm d_2$ and fiddling the signs a bit so that the cross-products cancel.

For the cross-products to cancel, we will need $b_1=-b_2$, $c_1=-c_2$, $a_1=a_2$, and $d_1=d_2$ or $b_1=b_2$, $c_1=c_2$, $a_1=-a_2$, and $d_1=-d_2$

We then also need $a_1a_2+2b_1b_2+3c_1c_2+6d_1d_2=1$. Given the above constraint, we therefore are looking for $a^2+6d^2-2b^2-3c^2=\pm1$ for some non-negative integers $a,b,c,d$. Also, we will need $ad-bd=\pm1$ so that the co-efficient of $\sqrt{6}$ is 2.

I'm sure there's a clever way to search for such a diophantine, but Python

>>> from itertools import product
>>> for a,b,c,d in product(range(5),repeat=4):
        if a*d-b*c in [-1,1] and a*a+6*d*d-2*b*b-3*c*c in [-1,1]:

0 1 1 1
1 2 0 1
2 1 1 0
2 1 3 2
3 1 2 1
4 3 1 1

The first solution I gave corresponds to the first line (and a bit of fiddling around with the signs). The second corresponds to the fourth and fifth. Obviously, there are many more...

a = $\frac{\sqrt[4]{23}}{4}$
b = $\frac{\sqrt[4]{23}}{4 \times \sqrt{2}}$
c = $\frac{\sqrt[4]{23}}{4 \times \sqrt{3}}$
d = $\frac{\sqrt[4]{23}}{4 \times \sqrt{6}}$
for all numbers.

We get

A = B = C = D =
$a + b \times \sqrt{2} + c \times \sqrt{3} + d \times \sqrt{6}$ =
$\frac{\sqrt[4]{23}}{4}$ + $\frac{\sqrt[4]{23}}{4}$ + $\frac{\sqrt[4]{23}}{4}$ + $\frac{\sqrt[4]{23}}{4}$ =
$4 \times \frac{\sqrt[4]{23}}{4}$ =


$A \times B \times C \times D = \sqrt[4]{23}^4$ = 23

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