Can Both Solutions of This Quadratic Equation Be Perfect Squares?

[anemone]

Here is this week's POTW:

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Let $a,\,b,\,c,\,d \in \mathbb{N}$ such that the equation $x^2-(a^2+b^2+c^2+d^2+1)x+ab+bc+cd+da=0$ has an integer solution. Prove that the other solution is integer too and both solutions are perfect squares.

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[anemone]

No one answered POTW #484. However, you can see the suggested solution as follows:

Spoiler

Let $x_1 \le x_2$ are roots. Then we have $x_1+x_2=1+a^2+b^2+c^2+d^2$, therefore if one of the roots is an integer, so the second root is an integer too.

We also have $x_1x_2=ab+bc+cd+da$.

Since $a^2+b^2+c^2+d^2=ab+bc+cd+da+\dfrac{(a-b)^2+(b-c)^2+(c-d)^2+(d-a)^2}{2}$, we get $x_1=1$, $a=b=c=d$ and so $x_2=(2a)^2$.