ChallengeQuadratic equation that has two roots

anemone

Paris la ville de l'amour
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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.

MarkFL

MarkFL

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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:

$$\displaystyle k\in\{1989,1993\}$$

anemone

anemone

Paris la ville de l'amour
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Very good Mark! And thanks for participating in this challenge problem!

MarkFL