Why three-digit integers only? (Part-2)

Since $A, B-A$ and $B$ are all integer squares, their square roots form a Pythagorean triple. Thus, the easiest place to look for numbers meeting the stated conditions is in the multiples of the $(3; 4; 5)$ triple. So, letting $A = 9t^2$ and $B = 25t^2$ for an integer $t$ , we see that $B-A = 16t^2$ ,

$A \text{ \# } B = 1000A + B = 9000t^2 + 25t^2 = 9025t^2 = (95t)^2$ , and

$\dfrac{A \text{ \# } B}{B} = \dfrac{(95t)^2}{(5t)^2} = 19^2$ .

Any choice of $t$ would yield $A$ and $B$ satisfying all the other conditions, but in order to make $A$ and $B$ three-digit numbers, $t$ would have to be $4, 5$ or $6$ , yielding values of ( for $(A,B)$ ):

$(144, 400); \quad (225,625), \quad (324,900)$

These are in fact the only possibilities. Suppose $A = x^2, B-A = y^2 \text{ and } B = z^2$ satisfy the given conditions. Let $q = gcd(x; y; z), x' = \frac{x}{q}, y' = \frac{y}{q} \text { and } z' = \frac{z}{q}$ .

Then $x'^2+ y'^2 = z'^2$ and $z'^2$ divides (A \text{ \# } B - B)/q^2 = 1000x'^2 . Since $x'$ and $z'$ are relatively prime, this means that $z'^2$ must divide $1000$ , so $z'$ must be $1, 2, 5$ or $10$ . By inspection, none of the others are possible values, so $z'$ must be $5$ . We can also rule out the remaining case of $x' = 4, y' = 3$ and $z' = 5$ [that is, $A = 16t^2$ and $B = 25t^2$ ] since $(A \text{ \# }B)/t^2 = 16025$ is not a perfect square.

So, adding the values of all possible values of $A, B$ , we get: $144 + 225 + 324 + 400 + 625 + 900 = \boxed{2618}$ .

If $A=B$ , then $\frac{A \text{\#}B}{B}=\frac{A \text{\#}A}{A}=\frac{1001A}{A}=1001$ will not be a square. Therefore $B>A$ . Let $B=b^2$ and $A=a^2$ , where $10\le a<b\le 31$ .

Note that if $\frac{A \text{\#}B}{B}$ is a square, then $B \frac{A \text{\#}B}{B}=A \text{\#}B$ will be a square too. So we just need to check one of them. Hence problem reduces to finding pairs of $(a, b)$ , where both of $b^2-a^2$ and $\frac{a^2 \text{\#}b^2}{b^2}$ are squares. Let $a=md$ and $b=nd$ , where $\gcd(a, b)=d$ . We get $\frac{a^2 \text{\#}b^2}{b^2}=1000 \frac{m^2}{n^2}+1=k^2$ It follows that $n^2 |1000$ because $\gcd(m, n)=1$ . So only possible values for $n$ will be $2, 5, 10$ . Since $m<n$ and $\gcd(m, n)=1$ we have a few cases to check. Only pair of $(m, n)$ that makes $1000 \frac{m^2}{n^2}+1$ square is $(3, 5)$ .

Now $4 \le d \le 6$ and there are three pairs of $(a=md, b=nd)$ to check. all three of them makes both of $b^2-a^2$ and $\frac{a^2 \text{\#}b^2}{b^2}$ squares and they are $(a, b)=(12, 20), (15, 25), (18, 30)$ The answer will be $12^2+20^2+15^2+25^2+18^2+30^2=2618$