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Problem Of The Week #414 Apr 26th, 2020

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
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 that both solutions are perfect squares.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to Opalg for his correct solution :cool: , which you can find below:

The sum of the two solutions is $a^2+b^2+c^2+d^2+1$, which is an integer. So if one of the solutions is an integer then obviously the other one must also be an integer. Also, the sum and the product of the solutions are positive, so the solutions must both be positive.

Next, $ab \leqslant \tfrac12(a^2+b^2)$ (with equality only if $a=b$), and there are similar inequalities for $bc$, $cd$ and $da$. Therefore $ab+bc+cd+da \leqslant a^2+b^2+c^2+d^2$, with equality only if $a=b=c=d$.

Let $\sigma = ab+bc+cd+da$. Then $a^2+b^2+c^2+d^2 + 1 \geqslant \sigma+1$. So the two solutions of the quadratic equation are positive integers with product $\sigma$ and sum at least $\sigma+1$. That can only happen if the integers are $1$ and $\sigma$ (because if two integers have a given product then their sum is greatest when their difference is greatest). So the sum of the roots is equal to $\sigma + 1$, in other words equality occurs in the above inequality. Therefore $a=b=c=d$, and the quadratic equation becomes $x^2 - (4a^2+1)x + 4a^2$, with solutions $1$ and $4a^2$. Since $1=1^2$ and $4a^2 = (2a)^2$, both solutions are perfect squares.
 
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