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

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Prove: if px2-(p+q)x+p=0&amp;nbsp;has an integer root,&amp;nbsp;px2+qx+p2-q=0&amp;nbsp;has an integer root too, for com&#039;&amp;nbsp;p,q∈N

prstenast4yi · Accepted Answer

Step 1In this answer, p and q are co&#039; integers such that q≥0 (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+p2−q=0If x=t is an integer root of px2−(p+q)x+p=0, thent(p+q−pt)=p.If t=0, then p=0. Because gcd(p,q)=1, this means q=1. Then, the equation px2+qx+p2−q=0 which is equivalent to x−1=0, has an integer solution x=1From now on, suppose that t≠0. That is, p=st for some integer s. Hence,p+q−pt=pt=s.Therefore,q=s+p(t−1)=s+st(t−1)=s(t2−t+1).Therefore, s divides p and q, which implies s=±1 as p  and q are co&#039;. Since q≥0, we get s=1Therefore, p=t and q=t2−t+1. Hence,px2+qx+p2−q=tx2+(t2−t+1)x+(t2−(t2−t+1))That is, x=1−t is an integer solution to the quation px2+qx+p2−q=0The condition q≥0 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+p2−q=0 does not have an integer solution.

hadnya1qd · Accepted Answer

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-&#039;, 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-&#039;, 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.

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

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