Problem 1 of the HMO 2025

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Problem 1 of the HMO 2025

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#1 h Feb 22, 2025, 12:32 PM

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Let $P(x)=x^4+5x^3+mx^2+5nx+4$ have $2$ distinct real roots, which sum up to $-5$ . If $m,n \in \mathbb {Z_+}$ , find the values of $m,n$ and their corresponding roots.

This post has been edited 1 time. Last edited by GreekIdiot, Feb 22, 2025, 3:33 PM

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#2 h Feb 22, 2025, 1:33 PM

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$x^4+5x^3+mx^2+nx+4=(x^2+5x+a)(x^2+b)$ and $25 - 4a=d^2$ for some $d$

$a+b=m$ , $5b=n$ and $ab=4$

So $a|4 \to a$ can be $-4,-2,-1,1,2,4$
As $25-4a$ is square then $a=4$ and roots are $-1,-4$
$b=1,n=5,m=5$ and $P(x)=x^4+5x^3+5x^2+5x+4$

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#3 h Feb 22, 2025, 1:43 PM

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GreekIdiot wrote:

Let $P(x)=x^4+5x^3+mx^2+nx+4$ have $2$ distinct real roots, which sum up to $-5$ . If $m,n \in \mathbb {Z_+}$ , find the values of $m,n$ and their corresponding roots.

Sum of four real or complex roots is $-5$ , same as sum of the two distinct real roots. So the sum of the two others is zero and we have ;
$x^4+5x^3+mx^2+nx+4=(x^2+5x+a)(x^2+b)$ for some $a<\frac{25}4$ (in order first quadratic has two distinct real roots)

This gives $ab=4$ , $5b=n$ and $a+b=m$

SO $b=\frac n5$ , $a=m-\frac n5$ and $\frac n5(m-\frac n5)=4$ , which is $n(5m-n)=100$

This implies $5|n$ and so $b=\frac n5\in\mathbb Z_{>0}$ and so $a\in\mathbb Z$ and $ab=4$ implies $(a,b)\in\{(1,4),(2,2),(4,1)\}$

And so three solutions ;
$\boxed{\text{S1 : }(m,n)=(5,20)\text{ with real roots }\frac{-5\pm \sqrt {21}}2}$

$\boxed{\text{S2 : }(m,n)=(4,10)\text{ with real roots }\frac{-5\pm 2\sqrt {17}}2}$

$\boxed{\text{S3 : }(m,n)=(5,5)\text{ with real roots }\frac{-5\pm 3}2}$

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#4 h Feb 22, 2025, 3:34 PM

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guys I edited the post I made a typo. but the solution is very similar

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#5 h Apr 6, 2025, 2:20 PM

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pco wrote:

GreekIdiot wrote:

Let $P(x)=x^4+5x^3+mx^2+nx+4$ have $2$ distinct real roots, which sum up to $-5$ . If $m,n \in \mathbb {Z_+}$ , find the values of $m,n$ and their corresponding roots.

Sum of four real or complex roots is $-5$ , same as sum of the two distinct real roots. So the sum of the two others is zero and we have ;
$x^4+5x^3+mx^2+nx+4=(x^2+5x+a)(x^2+b)$ for some $a<\frac{25}4$ (in order first quadratic has two distinct real roots)

This gives $ab=4$ , $5b=n$ and $a+b=m$

SO $b=\frac n5$ , $a=m-\frac n5$ and $\frac n5(m-\frac n5)=4$ , which is $n(5m-n)=100$

This implies $5|n$ and so $b=\frac n5\in\mathbb Z_{>0}$ and so $a\in\mathbb Z$ and $ab=4$ implies $(a,b)\in\{(1,4),(2,2),(4,1)\}$

And so three solutions ;
$\boxed{\text{S1 : }(m,n)=(5,20)\text{ with real roots }\frac{-5\pm \sqrt {21}}2}$

$\boxed{\text{S2 : }(m,n)=(4,10)\text{ with real roots }\frac{-5\pm 2\sqrt {17}}2}$

$\boxed{\text{S3 : }(m,n)=(5,5)\text{ with real roots }\frac{-5\pm 3}2}$

Why (m,n)=(5,20) and (4,10)??

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#6 h Apr 8, 2025, 2:08 PM

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eric201291 wrote:

Why (m,n)=(5,20) and (4,10)??

Because we got three cases for $(a,b)$ : $(1,4),(2,2),(4,1)$
And we had $(m,n)=(a+b,5b)$ and so $(m,n)$ is $(5,20),(4,10),(5,5)$

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#7 h Apr 16, 2025, 2:12 PM

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thanks
pco

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