hapantad2j

2022-04-29

Prove: if p${x}^{2}-(p+q)x+p=0$ has an integer root, $p{x}^{2}+qx+{p}^{2}-q=0$ has an integer root too, for com' $p,q\in \mathbb{N}$

prstenast4yi

Beginner2022-04-30Added 12 answers

Step 1

In this answer, p and q are co' integers such that$q\ge 0$ (and p can be zero or negative). We want to show that if $p{x}^{2}-(p+q)x+p=0$ has an integer solution, then so does $p{x}^{2}+qx+{p}^{2}-q=0$

If$x=t$ is an integer root of $p{x}^{2}-(p+q)x+p=0$ , then

$t(p+q-pt)=p.$

If$t=0$ , then $p=0$ . Because $gcd(p,q)=1$ , this means $q=1$ . Then, the equation $p{x}^{2}+qx+{p}^{2}-q=0$ which is equivalent to $x-1=0$ , has an integer solution $x=1$

From now on, suppose that$t\ne 0$ . That is, $p=st$ for some integer s. Hence,

$p+q-pt=\frac{p}{t}=s.$

Therefore,

$q=s+p(t-1)=s+st(t-1)=s({t}^{2}-t+1).$

Therefore, s divides p and q, which implies$s=\pm 1$ as p and q are co'. Since $q\ge 0$ , we get $s=1$

Therefore,$p=t$ and $q={t}^{2}-t+1$ . Hence,

$p{x}^{2}+qx+{p}^{2}-q=t{x}^{2}+({t}^{2}-t+1)x+({t}^{2}-({t}^{2}-t+1))$

That is,$x=1-t$ is an integer solution to the quation $p{x}^{2}+qx+{p}^{2}-q=0$

The condition$q\ge 0$ is necessary. Note that $p=-2$ and $q=-3$ satisfies the condition that $p{x}^{2}-(p+q)x+p=0$ has an integer solution, but $p{x}^{2}+qx+{p}^{2}-q=0$ does not have an integer solution.

In this answer, p and q are co' integers such that

If

If

From now on, suppose that

Therefore,

Therefore, s divides p and q, which implies

Therefore,

That is,

The condition

hadnya1qd

Beginner2022-05-01Added 15 answers

From the first equation, the product of the roots is 1. So they are both 1 or both -1. If they are both 1, substituting $x=1$ into the first equation gives $q=p$ . Since p and q are co-', $p=1,q=1$ or $p=-1,q=-1$ If they are both -1, substituting $x=-1$ into the first equatio gives $q=-3p$ . Since p and q are co-', $p=1,q=-3$ or $p=-1,q=3$ In all four cases for p and q, substituting the values for p and q into the second equation gives solutions for x that are integers.

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