Numbers are beautiful and extremely charming, just as—
There is a type of number that, when
split into two numbers, the square of the sum of these
two numbers equals the original
number. For example: \begin{aligned}
2025&=(20+25)^2 \\
88209&=(88+209)^2 \\
152344237969&=(152344+237969)^2 \\
&...
\end{aligned}
Below are some studies regarding this type of number:
The essence of these numbers is: (A+B)^2=10^nA+B. For the general form (A+B)^2=kA+B, we have: A = \frac{k}{2} - B \pm \sqrt{\frac{k^2}{4} - (k-1)B} Therefore, generally, for a suitable B, two corresponding values of A can be found.
For the case (A+B)^2=10^{2n}A+B, A and B can be: B = \frac{10^{2n}}{4}, \quad A = \frac{10^{2n}}{4} \pm \frac{10^n}{2}
General Solution:
For (A+B)^2=10^nA+B, we have (A+B)(A+B-1)=(10^n-1)A. Regarding this, we know that we need to find the product of two adjacent natural numbers that is a multiple of (10^n-1). Suppose we let 10^n-1=X \cdot Y and A=M \cdot N, and satisfy M \cdot X = N \cdot Y \pm 1.
X and Y can be pre-determined from 10^n-1. By finding a suitable N, we can determine M. Thus:
In the form M \cdot X = N \cdot Y + 1, we have A=MN and A+B=NY+1;
In the form M \cdot X = N \cdot Y - 1, we have A=MN and A+B=NY.
In this process, the Remainder Theorem and factorization will be used.
Examples:
3.1 Finding 2-digit A and B,
That is, (A+B)(A+B-1)=99A.
Let 99=9 \cdot 11 (It cannot be split
into 3 \cdot 33 or 1 \cdot 99; think about why?). In the form
M \cdot X = N \cdot Y - 1:
We have (11N-1) \pmod 9 = 0.
Thus N is a number of the form 5+9p. Taking N=5, we get M=6, A=30, and B=25. At the same time, for B=25, another A is 20. That is: \begin{aligned} (30+25)^2 &= 3025 \\ (20+25)^2 &= 2025 \end{aligned}
3.2 Finding 6-digit A and B,
That is, (A+B)(A+B-1)=(10^6-1)A.
Since 10^6-1=3^3 \cdot 7 \cdot 11 \cdot 13
\cdot 37, we can let 10^6-1=143 \cdot
6993 (not unique). In the form M \cdot
X = N \cdot Y - 1:
We have (6993N-1) \pmod{143} = 0.
We can solve for N as a number of the form 51+143p. Taking N=51, then M=2494, A=127194, and B=229449. At the same time, for B=229449, another A is 413908. That is: \begin{aligned} (127194+229449)^2 &= 127194229449 \\ (413908+229449)^2 &= 413908229449 \end{aligned}
At this point, the problem is basically solved. Following this line of thought, we can find more such square numbers. Moreover, in general, the number of such square numbers is infinite.
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