A number with a special property

$\begin{aligned} \overline{abc} & = \overline{ab}^2 - c^2 \\ 100a + 10b + c & = (10a+b)^2 - c^2 \\ \blue{100a} + 10b + c & = \blue{100a^2}+20ab + b^2 - c^2 \end{aligned}$

It is obvious from the above that $a=1$ . Then the equation becomes:

$\begin{aligned} 10 b + c & = 20b + b^2 - c^2 \\ c^2 + c & = b^2 + 10b \\ c^2 + c + 25 & = (b+5)^2 \end{aligned}$

This means that $c^2 + c + 25$ must be a perfect square. From $0$ to $9$ , $c^2 + c + 25$ is a perfect square when $c=0$ and $c=7$ and we have:

$\begin{cases} c = 0 & c^2 + c + 25 = 5^2 & \implies b + 5 = 5 & \red{\implies b = 0 = c} & \small \red{\text{Solution invalid}} \\ c = 7 & c^2 + c + 25 = 9^2 & \implies b + 5 = 9 & \implies b = 4 \end{cases}$

Therefore $\overline{abc} = \boxed{147}$ .

Let $\overline{ab} = x$ , where $x$ is a natural number satisfying $10 \leq x \leq 99$ .

We have: $\overline{abc} = 10\overline{ab}+c = 10x+c$

Therefore, $10x+c=x^2-c^2 \Leftrightarrow c(c+1) = x(x-10)$ $(*)$

As $c \leq 9$ , $c(c+1) \leq 90 \Leftrightarrow x(x-10) \leq 90 \Leftrightarrow x \leq 15$ .

Let's test all cases of $x \in \{ 10,11,12,13,14,15 \}$ by checking if $x(x+10)$ can be written as the product of two consecutive numbers. We can see that $x \in \{ 10,14 \}$ is the only solution.

Substituting the results in $(*)$ to solve for $c$ , we can conclude that $\overline{abc}$ is either $100$ or $147$ .

As $a,b,c$ are different digits, the only solution is $\boxed{147}$ .

\( \begin{align} \overline{abc} &= \overline{ab}^2 - c^2 \\ 100a+10b+c &= (10a+b)^2-c^2 \\ 100a+10b+c &= 100a^2 + 20ab + b^2 -c^2 \\ \\ \implies c^2-b^2 + c &= 100a(a-1) + 10b (2a-1) \hspace{5mm} \text{ LHS max value }= 81 - 0 + 9 < 100 \therefore \boxed{a=1} \\ \implies c^2-b^2 + c &= 10b \\ c ( c +1 ) &= b (10+ b) \hspace{30mm} \text{ either one of } c, (c+1) \text{ is even } \rightarrow b >0 \text{ is even } \\ \\ \end{align} \\

\left\{ \begin{array} &b=2 &c(c+1)&= 24 &= 4\cdot 6 &\implies &\text{no solution}\\ &\boxed{b=4} &c(c+1)&= 56 &= 7\cdot 8 &\implies &\boxed{c=7}\\ &b\geq6 &c(c+1)&\geq96 &\geq9\cdot 10 &\implies &\text{no solution } \\ \end{array} \right. \\ \\ \hspace{1mm} \\

\\ \therefore \overline{abc} = \boxed{147} \)