Challenge Quadratic equation that has two roots

anemone

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#1
Find all integer valuse of \(\displaystyle k\) such that the quadratic expression \(\displaystyle (x+k)(x+1991)+1\) can be factored as a product of \(\displaystyle (x+p)(x+q)\), where \(\displaystyle p\) and \(\displaystyle q\) are integers.
 
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MarkFL

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#2
If we expand and equate the two expressions, we obtain:

\(\displaystyle x^2+(k+1991)x+1991k+1=x^2+(p+q)x+pq\)

Equating coefficients, we get:

\(\displaystyle k+1991=p+q\)

\(\displaystyle 1991k+1=pq\)

Now, the second equation implies:

\(\displaystyle q=\frac{1991k+1}{p}\)

Substituting into the first equation, we obtain:

\(\displaystyle k+1991=p+\frac{1991k+1}{p}\)

Multiply by \(p\) and arrange in standard form:

\(\displaystyle p^2-(k+1991)p+1991k+1=0\)

One condition we require for the roots of this quadratic to be an integer is for \(k\) to be odd, so let:

\(\displaystyle k=2r+1\) where \(r\in\mathbb{Z}\)

And we have:

\(\displaystyle p^2-(2r+1+1991)p+1991(2r+1)+1=0\)

Or:

\(\displaystyle p^2-2(r+996)p+2(1991r+996)=0\)

Now, we also require the discriminant to be a perfect square:

\(\displaystyle (-2(r+996))^2-4(1)2(1991r+996)=2^2\left((r+996)^2-2(1991r+996)\right)=2^2(r-994)(r-996)\)

Let's let:

\(\displaystyle s=r-996\)

And look at the following product as a perfect square:

\(\displaystyle s(s+2)=n^2\)

\(\displaystyle s^2+2s-n^2=0\)

The discriminant here is:

\(\displaystyle 4+4n^2=4(n^2+1)\)

The only two perfect squares that differ by 1 are 0 and 1, and so we must have \(n=0\). This implies:

\(\displaystyle s=0\implies r=996\)

\(\displaystyle s=-2\implies r=994\)

Now, these together imply:

\(\displaystyle k\in\{1989,1993\}\)
 
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