Find the perfect squares of four digits whose square root is the sum of the numbers obtained if we separate the first two digits from the last two

Question

the question

We are asking for the perfect squares of four digits whose square root is the sum of the numbers obtained if we separate the first two digits from the last two

the idea

let the number be $\overline{abcd}$, where a,b,c,d are digits

$32^2\leq \overline{abcd} \leq 99^2$, which means $32\leq \sqrt{\overline{abcd} } \leq 99$

I don't really understand the part with 'sum'... there are 2 or more numbers to add? Hope one of you can help me! Thank you!

Answer 1

It seems that if there is an algorithm to find striaghtforwardly the solution it is difficult to find out it. So we try to reduce the $68$ possibilities to a smaller number.

Let $X, A, B$ integers such that $$\begin{cases}X^2=100A+B\\X=A+B\end{cases}\Rightarrow A=\frac{X(X-1)}{99}$$ then, because of the factor $11$, the integer $A$ is among the following fourteen possibilities: $$\frac{33\times32}{99}\space\space\text { or }\space\space\frac{34\times33}{99}\\\frac{44\times43}{99}\space\space\text { or }\space\space\frac{45\times44}{99}\\\frac{55\times54}{99}\space\space\text { or }\space\space\frac{56\times55}{99}\\\frac{66\times65}{99}\space\space\text { or }\space\space\frac{67\times66}{99}\\\frac{77\times76}{99}\space\space\text { or }\space\space\frac{78\times77}{99}\\\frac{88\times87}{99}\space\space\text { or }\space\space\frac{89\times88}{99}\\\frac{99\times98}{99}\space\space\text { or }\space\space\frac{100\times99}{99}$$ Only $\dfrac{45\times44}{99}=20,\dfrac{55\times54}{99}=30,\dfrac{99\times98}{99}=98$ are integers as desired, of two digits (the other eleven quantities are fractional excepting the last equal to $100$). Inside of them are explicitly listed the only three solutions $\boxed{X=45,55,99}$

Answer 2

A necessary condition is to find positive integer solutions to $(x+y)^2 = 100x +y$, subject to $32 \leq x+y \leq 99$.

Writing this as $ (x+y)(x+y-1) = 99 x $, since $ \gcd(x+y, x+y - 1) = 1$, conclude that we must have:

• $x+y \equiv 0, 1 \pmod{9}$
• $x+y \equiv 0, 1 \pmod {11}$

Thus, we get that
$ x+y \equiv 0, 1, 45, 55 \pmod{99}$.

Finally, verify that $x+y = 45, 55, 99$ satisfy the conditions.
Note: In the verification, we have to show that $ x, y$ are integers from 0 to 100. In particular, this implies that $\lfloor \frac{ (x+y)^2}{100} \rfloor = x$.