Perfect Squares Only

Since $a$ , $b$ , $c$ are all perfect square numbers, we can rewrite them as squared numbers for some positive integers $s$ , $t$ , $u$ below:

$a = s^2; c = t^2; b = u^2$

Now from $b$ = $a$ + $c$ , $u^2$ = $s^2$ + $t^2$ .

That means the integers $s$ , $t$ , $u$ form a Pythagorean triple, and when we evaluate the quadratic discriminant, it will result in absolute difference between two squares:

$\sqrt{b^2 - 4ac}$ = $\sqrt{(a + c)^2 - 4ac}$ = $\sqrt{a^2 - 2ac + c^2}$ = $\sqrt{(a - c)^2 }$ = $\mid{a - c}\mid$

Substituting $a$ = $s^2$ ; $c$ = $t^2$ , we will get:

$\mid{a - c}\mid$ = $\mid{s^2-t^2}\mid$ = $\mid{s - t}\mid$ $\mid{s + t}\mid$

Since the condition restricts that the discriminant is prime, that means $\mid{s - t}\mid$ = 1. Otherwise, the product will be a composite number.

Then in case $s$ = $t$ + 1; $u^2$ = $(t+1)^2$ + $t^2$ = $2t^2$ + 2 $t$ + 1.

Clearly, $u$ is odd and equals to 2 $n$ + 1 for some integer $n$ .

$(2n+1)^2$ = $2t^2$ + 2 $t$ + 1

$4n^2$ + 4 $n$ = $2t^2$ + 2 $t$

$2n^2$ + 2 $n$ = $t^2$ + $t$

$2n$ [ $n+1$ ] = $t$ [ $t+1$ ]

This product is a multiplication of 2 consecutive natural numbers, which is also a multiplication of 2 consecutive numbers when it's divided by 2. Of all the consecutive pairs, only $3\times 4$ = $2$ [ $2\times 3$ ] fits the constraint. (For $t$ = $s$ + 1, it will have similar results.)

In other words, (3 , 4 , 5) and (4 , 3 , 5) are the only applicable triples for (s , t , u). Hence, the discriminant will equal to $\sqrt{25^2 - 4(9\times 16)}$ = $7$ .

Nonetheless, if $a$ = $4^2$ ; $c$ = $3^2$ , the roots will equal to $\dfrac{(25 + 7)}{32}$ = $\dfrac{32}{32}$ = $1$ or $\dfrac{(25 - 7)}{32}$ = $\dfrac{18}{32}$ , which never exceeds 1. Therefore, $a$ = $3^2$ ; $c$ = $4^2$ .

Hence, $d$ = $\dfrac{(25 + 7)}{18}$ = $\dfrac{32}{18}$ = $\dfrac{16}{9}$ = $\dfrac{c}{a}$ .

Therefore, $\sqrt{abcd}$ = $\sqrt{9\times 25\times 16\times (\frac{16}{9})}$ = $16\times 5$ = $80$ .