Prove: if px^2 - (p+q)x + p = 0ZK has

hapantad2j

hapantad2j

Answered question

2022-04-29

Prove: if px2-(p+q)x+p=0 has an integer root, px2+qx+p2-q=0 has an integer root too, for com' p,qN

Answer & Explanation

prstenast4yi

prstenast4yi

Beginner2022-04-30Added 12 answers

Step 1
In this answer, p and q are co' integers such that q0 (and p can be zero or negative). We want to show that if px2(p+q)x+p=0 has an integer solution, then so does px2+qx+p2q=0
If x=t is an integer root of px2(p+q)x+p=0, then
t(p+qpt)=p.
If t=0, then p=0. Because gcd(p,q)=1, this means q=1. Then, the equation px2+qx+p2q=0 which is equivalent to x1=0, has an integer solution x=1
From now on, suppose that t0. That is, p=st for some integer s. Hence,
p+qpt=pt=s.
Therefore,
q=s+p(t1)=s+st(t1)=s(t2t+1).
Therefore, s divides p and q, which implies s=±1 as p and q are co'. Since q0, we get s=1
Therefore, p=t and q=t2t+1. Hence,
px2+qx+p2q=tx2+(t2t+1)x+(t2(t2t+1))
That is, x=1t is an integer solution to the quation px2+qx+p2q=0
The condition q0 is necessary. Note that p=2 and q=3 satisfies the condition that px2(p+q)x+p=0 has an integer solution, but px2+qx+p2q=0 does not have an integer solution.
hadnya1qd

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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