Triple of integers

Let the three positive integers be (respectively) $a,b,c$ . From the question, we have $b^2=a+c=c^2-a^2$

Since $a,c$ are positive, we can divide by $a+c$ to find that $c-a=1$

So we're looking for Pythagorean triples with the hypotenuse one more than the length of one of the legs. The obvious candidate - $3,4,5$ - works, with $a=4$ , $b=3$ , $c=5$ giving an answer of $\boxed{5}$ .

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We can find all solutions: since $c-a=1$ , we have $b^2=2a+1$ . So $b$ must be odd, and greater than $1$ ; if we let $b=2t+1$ , then $a=2t^2+2t, \;\;\; b=2t+1, \;\;\; c=2t^2+2t+1$

for positive integer $t$ .

Let the three positive integers be $a$ , $b$ , and $c$ . Then we have:

$\begin{cases} a + c = b^2 & ...(1) \\ a^2 + b^2 = c^2 & ...(2) \end{cases}$

From $(2)$ , since $a+c = b^2$ :

$\begin{aligned} a^2 + a + c & = c^2 \\ a+c & = c^2 - a^2 \\ & = (c+a)(c-a) \\ c-a & = 1 \\ \implies c & = a + 1 \end{aligned}$

From $(1)$ , since $c=a+1$ : $a+c = 2a+1 = b^2$ . This means that $b$ is odd. The smallest odd perfect square is $1$ , then $a=0$ , which is unacceptable. The next smallest odd perfect square is $3^2$ , then $a=4$ and $c=4+1=5$ .

Therefore $(a,b,c) = (4,3,5)$ and $\dfrac {abc}{a+b+c} = \dfrac {4\times 3 \times 5}{4+3+5} = \boxed 5$ .