MHB Can Both Solutions of This Quadratic Equation Be Perfect Squares?

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The discussion revolves around a problem of the week (POTW) involving a quadratic equation with natural number parameters. The equation is structured such that it has an integer solution, and the challenge is to prove that the other solution is also an integer and that both solutions are perfect squares. Despite the lack of responses to the previous POTW, a suggested solution is provided for this current problem. The focus is on demonstrating the properties of the solutions derived from the given quadratic equation. The mathematical exploration emphasizes the relationship between the coefficients and the nature of the roots.
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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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No one answered POTW #484. However, you can see the suggested solution as follows:

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 $x_2=(2a)^2$.
 
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