{"id":4498,"date":"2025-09-20T15:59:00","date_gmt":"2025-09-20T19:59:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4498"},"modified":"2025-10-05T23:38:31","modified_gmt":"2025-10-06T03:38:31","slug":"usamts-5-2-36","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4498","title":{"rendered":"USAMTS 5\/2\/36"},"content":{"rendered":"\n

Prove that there is no polynomial $P(x)$ with integer coefficients such that $$
P(\\sqrt[3]{5}+\\sqrt[3]{25})=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$\ud83d\udd11<\/a><\/p>\n

<\/p>\n\n\n\n

Proof:<\/strong> Assume that there is a polynomial $P(x)$ of degree $n$, with integer coefficients, and without trivial root of $0$, such that
$$P(\\sqrt[3]{5}+\\sqrt[3]{25})=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$
Therefore, $P(x)$ can be expressed as $$P(x)=a_nx^n+a_{n-1}x^{n-1}+\\ldots a_2x^2+a_1x+a_0$$
where $n\\ge 0$, $a_i$ are integers, $i=0,1,2,\\ldots, n$, and $a_0\\ne 0$, and $a_n\\ne 0$.

Let $t=\\sqrt[3]{5}+\\sqrt[3]{25}$, then $$P(t)=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$ Because $$t^3=(\\sqrt[3]{5}+\\sqrt[3]{25})^3=5+15(\\sqrt[3]{5}+\\sqrt[3]{25})+25=30+15t$$
Therefore the minimum polynomial equation with a root as $t=\\sqrt[3]{5}+\\sqrt[3]{25}$ is $$x^3-15x-30=0$$ By Euclidean division of polynomials with integer coefficients, for divisor $x^3-15x-30$, there exists a unqiue $Q(x)$ and a unique $R(x)$ so that
$$P(x)=(x^3-15x-30)Q(x)+R(x)$$
where $Q(x)$ is a polynomial of degree $n-3$ with integer coefficients, and $R(x)$ is a
polynomial of degree at most $2$ with integer coefficients. Note that if $n<3, Q(x)=0$.

Because $t=\\sqrt[3]{5}+\\sqrt[3]{25}$ is a root of polynomial equation $x^3-15x-30=0$, we have $$P(t)=(t^2-15t-30)Q(t)+R(t)=0\\cdot Q(t)+R(t)=R(t)=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$ Let $$R(x)=ax^2+bx+c$$ where $a$, $b$, and $c$ are integers, which also covers the cases when degree $n$ is $0$, $1$, or $2$, we have<\/p>\n\n\n\n

$$\\begin{array}{ccccl}
P(t)&=&R(t)&=&a(\\sqrt[3]{5}+\\sqrt[3]{25})^2+b(\\sqrt[3]{5}+\\sqrt[3]{25})+c \\\\
& & &=&a(\\sqrt[3]{25}+10+5\\sqrt[3]{5})+b(\\sqrt[3]{5}+\\sqrt[3]{25})+c \\\\
& & &=&(5a+b)\\sqrt[3]{5}+(a+b)\\sqrt[3]{25}+(10a+c) \\\\
& & &=&2\\sqrt[3]{5}+3\\sqrt[3]{25}
\\end{array}$$<\/p>\n\n\n\n

Because $\\sqrt[3]{5}$ and $\\sqrt[3]{25}$ are irrational numbers that do not share any rational common factors, in order to satisfy the following equation:<\/p>\n\n\n\n

$$(5a+b)\\sqrt[3]{5}+(a+b)\\sqrt[3]{25}+(10a+c)=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$ where $a+b$, $5a+b$, $10a+c$ are integers, which are rationals, we must have the following:<\/p>\n\n\n\n

$$
\\left\\{
\\begin{array}{rcrcrc@{\\qquad}l}
5a & + & b & & & = & 2 \\\\
a & + & b & & & = & 3 \\\\
10a & & & + & c & = & 0
\\end{array}
\\right.
$$<\/p>\n\n\n\n

Solving the above equations, we have $a=-\\dfrac{1}{4}$, $b=\\dfrac{13}{4}$, and $c=\\dfrac{5}{2}$. Hence,<\/p>\n\n\n\n

$$R(x)=-\\dfrac{1}{4}x^2+\\dfrac{13}{4}x+\\dfrac{5}{2}$$ However, the above result contradicts the assumption that $R(x)$ is a polynomial with integer coefficients, nor does it work if degree $n$ is $0$ or $1$. Additionally, because $$P(x)=(x^3-15x-30)Q(x)+R(x)$$ where $Q(x)$ is a polynomial of degree $n-3$ with integer coefficients, it also contradicts the assumption that $P(x)$ is a polynomial with integer coefficients.

Therefore, the original assumption is false, which completes the proof that there is no polynomial $P(x)$ with integer coefficients such that
$$P(\\sqrt[3]{5}+\\sqrt[3]{25})=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$$\\boxed{\\phantom{\\ }}$<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Prove that there is no polynomial $P(x)$ with integer coefficients such that $$P(\\sqrt[3]{5}+\\sqrt[3]{25})=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$\ud83d\udd11 Proof: Assume that there is a polynomial $P(x)$ of degree $n$, with integer coefficients, and without trivial root of $0$, such that$$P(\\sqrt[3]{5}+\\sqrt[3]{25})=2\\sqrt[3]{5}+3\\sqrt[3]{25}$$Therefore, $P(x)$ can be expressed as … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4498"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4498"}],"version-history":[{"count":49,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4498\/revisions"}],"predecessor-version":[{"id":5333,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4498\/revisions\/5333"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4498"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4498"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4498"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}