2 degree polynomial

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#1 h Mar 24, 2025, 6:13 AM

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Let $P_{1}(x)= x^2 +b_{1}x +c_{1}, ... , P_{n}(x)=x^2+ b_{n}x+c_{n}$ , $P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$ $\forall i,j ,1 \leq i<j \leq n : P_{i}(x) \ne P_{j}(x)$ .
$\forall i,j, 1\leq i<j \leq n : Q_{i,j}(x)= P_{i}(x) + P_{j}(x)$ polynomial with only one root.
$max(n)=?$

This post has been edited 8 times. Last edited by PrimeSol, Mar 24, 2025, 6:34 AM

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#3 h Mar 25, 2025, 2:45 AM

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Any solution?

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#4 h Tuesday at 9:02 AM • 1 Y

Y by PrimeSol

PrimeSol wrote:

Let $P_{1}(x)= x^2 +b_{1}x +c_{1}, ... , P_{n}(x)=x^2+ b_{n}x+c_{n}$ , $P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$ $\forall i,j ,1 \leq i<j \leq n : P_{i}(x) \ne P_{j}(x)$ .
$\forall i,j, 1\leq i<j \leq n : Q_{i,j}(x)= P_{i}(x) + P_{j}(x)$ polynomial with only one root.
$max(n)=?$

$n=3$ fits, for example with $(P_1,P_2,P_3)=(x^2+4x,x^2-4x,x^2+2)$
We'll show that $n=4$ does not fit.

Let us suppose all properties fit with some $n\ge 4$ :

Note that only possibility for $P_i+P_j$ in order to have only one root is $2(x+\alpha)^2$

Writing $P_1+P_2=2(x+a)^2$ , $P_1+P_3=2(x+b)^2$ and $P_2+P_3=2(x+c)^2$ , we get :
$P_1=(x+a)^2+(x+b)^2-(x+c)^2$ , $P_2=(x+a)^2-(x+b)^2+(x+c)^2$ and $P_3=-(x+a)^2+(x+b)^2+(x+c)^2$

Let us then consider $P_1,P_2,P_4$
$P_1+P_2=2(x+a)^2$ , $P_1+P_4=2(x+u)^2$ and $P_2+P_4=2(x+v)^2$ , we get :
$P_1=(x+a)^2+(x+u)^2-(x+v)^2$ , $P_2=(x+a)^2-(x+u)^2+(x+v)^2$ and $P_4=-(x+a)^2+(x+u)^2+(x+v)^2$

Looking at the two expressions of $P_1,P_2$ , we get $(x+b)^2-(x+c)^2=(x+u)^2-(x+v)^2$ $\forall x$ and so $b-c=u-v$ and $b^2-c^2=u^2-v^2$ and so :
Either $b=u$ and $c=v$ and then $P_3=P_4$ , impossible
Either $b=c$ and $u=v$ and then $P_1=P_2$ , impossible

And so no possibility for $n=4$ and $\boxed{\max n=3}$

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#5 h Yesterday at 4:20 AM

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pco wrote:

PrimeSol wrote:

Let $P_{1}(x)= x^2 +b_{1}x +c_{1}, ... , P_{n}(x)=x^2+ b_{n}x+c_{n}$ , $P_{i}(x)\in \mathbb{R}[x], \forall i=\overline{1,n}.$ $\forall i,j ,1 \leq i<j \leq n : P_{i}(x) \ne P_{j}(x)$ .
$\forall i,j, 1\leq i<j \leq n : Q_{i,j}(x)= P_{i}(x) + P_{j}(x)$ polynomial with only one root.
$max(n)=?$

$n=3$ fits, for example with $(P_1,P_2,P_3)=(x^2+4x,x^2-4x,x^2+2)$
We'll show that $n=4$ does not fit.

Let us suppose all properties fit with some $n\ge 4$ :

Note that only possibility for $P_i+P_j$ in order to have only one root is $2(x+\alpha)^2$

Writing $P_1+P_2=2(x+a)^2$ , $P_1+P_3=2(x+b)^2$ and $P_2+P_3=2(x+c)^2$ , we get :
$P_1=(x+a)^2+(x+b)^2-(x+c)^2$ , $P_2=(x+a)^2-(x+b)^2+(x+c)^2$ and $P_3=-(x+a)^2+(x+b)^2+(x+c)^2$

Let us then consider $P_1,P_2,P_4$
$P_1+P_2=2(x+a)^2$ , $P_1+P_4=2(x+u)^2$ and $P_2+P_4=2(x+v)^2$ , we get :
$P_1=(x+a)^2+(x+u)^2-(x+v)^2$ , $P_2=(x+a)^2-(x+u)^2+(x+v)^2$ and $P_4=-(x+a)^2+(x+u)^2+(x+v)^2$

Looking at the two expressions of $P_1,P_2$ , we get $(x+b)^2-(x+c)^2=(x+u)^2-(x+v)^2$ $\forall x$ and so $b-c=u-v$ and $b^2-c^2=u^2-v^2$ and so :
Either $b=u$ and $c=v$ and then $P_3=P_4$ , impossible
Either $b=c$ and $u=v$ and then $P_1=P_2$ , impossible

And so no possibility for $n=4$ and $\boxed{\max n=3}$

Thank you.

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