Radicands of square roots of the 2020s, written in simplest radical form

Question

As of the time of writing, the current decade is the 2020s. An interesting property of this decade is that there are 3 years that satisfy the property that the square-free part (https://oeis.org/A007913) of the year is a single-digit number, i.e. when $\sqrt{n}$ is expressed in simplest radical form, the radicand is less than 10.

The years' corresponding radicals are: $\sqrt{2023} = 17\sqrt7$, $\sqrt{2025} = 45$, and $\sqrt{2028} = 26\sqrt3$. (Squares have 1 as the radicand, because the $\sqrt{1}$ part is implicit, e.g. $\sqrt{2025}$ is also $45\sqrt{1}$).

I wondered: were there were finitely many decades that satisfy this property? To investigate, I wrote a simple Python 3 program; the below program finds all such decades less than $10^{12}$ where there are at least 3 years that satisfy the above property.

import bisect

LIMIT = 10**6

# 1, 2, 3, 5, 6, 7 are the possible single-digit radicands
values = sorted(a**2 * b for a in range(1, LIMIT) for b in [1, 2, 3, 5, 6, 7])

for i in range(len(values) - 2):
    if values[i+2] - values[i] < 10 and values[i] // 10 == values[i+2] // 10:
        if i == 0 or values[i] // 10 != values[i-1] // 10:
            last_idx = bisect.bisect_left(values, (values[i] // 10 + 1) * 10)
            print(values[i:last_idx])

This is the output:

[1, 2, 3, 4, 5, 6, 7, 8, 9]
[12, 16, 18]
[20, 24, 25, 27, 28]
[45, 48, 49]
[121, 125, 128]
[242, 243, 245]
[720, 722, 726, 729]
[841, 845, 847]
[2023, 2025, 2028]
[2880, 2883, 2888]
[4800, 4802, 4805]
[5041, 5043, 5046]
[6724, 6727, 6728]
[15123, 15125, 15129]

I was quite amazed that the previous decade with such a property was the 840s! We must be quite lucky, after all. In addition, it appears that the 15120s is the last decade with the above property.

What makes this particularly amazing also is that it is already known that there are infinitely many decades with at least 2 years with this property - just take any solution $(x,y)$ to the Pell's equation $x^2-2y^2=1$, $x^2-3y^2=1$, or whatever.

I was wondering if there's any intuition or formal proof that there are indeed finitely many such decades.

My intuition is as follows: such a decade would correspond to a positive-integer solution $(x,y,z)$ of the two Pell's equations

$$ax^2-b_1y^2=c_1$$ $$ax^2-b_2z^2=c_2$$

with $a,b_i$ squarefree and $0 < a, b_i < 10, -10 < c_i < 10, c_i\neq 0$. While each individual Pell's equation might have infinitely many solutions, there should be (I have not actually proved this) finitely many joint solutions to the 2 equations. For example, in the case of the 2020s, the 2 equations are

$$7x^2 - y^2 = -2$$ $$7x^2 - 3z^2 = -5$$

with solution $(x,y,z) = (17, 45, 26)$.

I'm still curious to know if there's any other intuition/formal proof behind this. Please feel free to post any partial results/intuition - I'd really like to get others' perspectives on this!

Answer 1

A simultaneous solution of $ax^2-c_1=b_1y^2$ and $ax^2-c_2=b_2z^2$ as demanded gives rise to an integral point on $b_1b_2T^2 = (ax^2-c_1)(ax^2-c_2)$, and since $a\ne 0$ and $c_1\ne c_2$, the degree-$4$ polynomial on the right is separable, meaning that this equation defines a curve of genus $1$. There are only finitely many integral points by Siegel's theorem, and since there were also only finitely many possible values of $a,b_1,b_2,c_1,c_2$, one has finiteness in total.

(Of course the gist of this argument is mentioned in some of the sources on simultaneous Pell equations given in the comments.)