Kexue.fm: An Interesting Kind of Square Number

科学空间：一种有趣的平方数 苏剑林 · 2009-07-08 · Mathematics square fun number

Numbers are beautiful, endlessly fascinating — as shown by the following:

There is a kind of number that can be split into two parts such that the square of the sum of these two parts equals the original number. For example:

$$\begin{aligned}2025=&(20+25)^2\\88209=&(88+209)^2\\152344237969=&(152344+237969)^2\\ &...\end{aligned}$$

Below is some investigation into this class of numbers:

1. The essence of such numbers is this: $(A+B)^2=10^nA+B$, and for $(A+B)^2=kA+B$, we have

$A=k/2-B\pm\sqrt{{k^2}/{4}-(k-1)B}$

So, in general, for a suitable B we can find a corresponding pair of A's. more

1. For the case of $(A+B)^2=10^{2n}A+B$, A and B can be

$B={10^{2n}}/4,A={10^{2n}}/4\pm{10^n}/2$

1. A general method of solution:

For $(A+B)^2=10^nA+B$, we have $(A+B)(A+B-1)=(10^n-1)A$. From this we see that we need to find two consecutive natural numbers whose product is a multiple of $(10^n-1)$. Suppose we let $10^n-1=X\cdot Y,A=M\cdot N$, and require that $M\cdot X=N\cdot Y+-1$.

$X,Y$ can be fixed in advance by $10^n-1$, and we then determine M by finding a suitable N. Thus, in the form $M\cdot X=N\cdot Y+1$, A = MN, A + B = NY + 1;

in the form $M\cdot X=N\cdot Y-1$, $A=MN，A+B=NY$. Along the way, this involves the remainder theorem, factorization, and so on.

For example:

3.1 Finding two-digit A and B,

that is, $(A+B)(A+B-1)=99A$.

Let $99=9\cdot 11$ (it cannot be split as $3\cdot 33,1\cdot 99$ — think about why). In the form $M\cdot X=N\cdot Y-1$,

we have $(11N-1)\mod 9=0$

So N takes the form 5 + 9p; taking N = 5 gives M = 6, A = 30, B = 25. Also, for B = 25, another possible A is 20. That is,

$$\begin{aligned}(30+25)^2=&3025\\ (20+25)^2=&2025\end{aligned}$$

3.2 Finding six-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 may let $10^6-1=143\cdot 6993$ (not unique). In the form $M\cdot X=N\cdot Y-1$,

we have $(6993N-1)\mod 143=0$

We can find that N takes the form 51 + 143p; taking N = 51 gives M = 2494, A = 127194, B = 229449. Also, for B = 229449, another possible A is 413908. That is,

$$\begin{aligned}(127194+229449)^2=&127194229449\\ (413908+229449)^2=&413908229449\end{aligned}$$

At this point the problem has essentially been solved. Following this line of reasoning, we can find even more such square numbers. And in general, the number of such squares is infinite.