Two Square Variables?

Relevant wiki: Quadratic Discriminant - Problem Solving

$\begin{aligned} f(x) & = (x+1991)(x+a) + 1 \\ & = x^2 + (1991+a)x+1991a + 1 \end{aligned}$

For $f(x)$ to be a perfect square, it must have the form of $f(x) = x^2 + 2kx + k^2$ , therefore,

$\begin{aligned} \implies \left(\frac{1991+a}{2}\right)^2 & = 1991a + 1 \\ a^2 + 3982a + 3964081 & = 7964a + 4 \\ a^2 -3982a + 3964077 & = 0 \\ (a-1989)(a-1993) & = 0 \end{aligned}$

The sum of all $a$ is $\boxed{3982}$ (Of course, Vieta's formula confirms it).

To make things easier, consider the polynomial $x'(x'+a') + 1$ instead, where $x' = x+1991, a' = a-1991$ . Note that $x'(x'+a') + 1 = (x+1991)(x+a) + 1$ by substituting. Because $(x+1991)(x+a) + 1$ is a perfect square for all integer $x$ , we also have $x'(x'+a') + 1$ to be a perfect square for all integer $x'-1991$ , which means $x'(x'+a') + 1$ is a perfect square for all integer $x'$ . For the rest of the following solution, assume that we're working with the polynomial $x(x+a) + 1$ instead of $(x+1991)(x+a) + 1$ ; we will substitute back at the end.

First, we claim that $a$ is an integer. This is easy. Substitute $x = 1$ to get that $a+2$ is a perfect square, and hence an integer. Since 2 is an integer, we have $a$ is also an integer.

Now, we can divide into cases.

Case 1 : $a = \pm 2$

This is the easy case. The polynomial factorizes to $(x \pm 1)^2$ , so for integer $x$ , we also have integer $x \pm 1$ , and hence $(x \pm 1)^2$ is a perfect square. So $a = \pm 2$ are solutions.

Case 2 : $a = -1, 0, 1$

All these three fail with $x = 2$ ; we have $x(x+a) + 1$ to be 3, 5, 7 respectively. So these three are not solutions.

Case 3 : $a > 2$ or $a < -2$

If $a < -2$ , then substituting $x = 1$ gives $a+2$ to be a perfect square. But $a+2 < 0$ , so it's impossible to be a perfect square. If $a > 2$ , substituting $x = -1$ gives a similar conclusion. Thus none of these are solutions.

Thus the only solutions are $a = -2, 2$ .

Back to the original problem, we found the solutions for $a'$ . Getting the solutions for $a$ is easy: $a = a'+1991 = 1989, 1993$ , which sum to $\boxed{3982}$ .

We can write it as $x^2+(a+1991)x+1991a+1$ and since its a perfect square, its discriminant equals $0$ giving $(a+1991)^2-4(1991a+1)=0 \Rightarrow a^2-3982a+1991^2-4=0$ . Since the question asks for sum, by vieta's formula we have sum as $\boxed{3982}$

This problem is quite simple to do analytically. If a quadratic is a perfect square then it must be of the form $(x+c)^2$ . In other words, the minimum of the quadratic must be at $0$ . Since $(x+1991)(x+a)+1$ is a quadratic expression, we must have its minimum at zero. Equivalent of this is $(x+1991)(x+a)$ has its minimum at $-1$ The minimum of a quadratic occurs halfway between it's roots which in this case must be at $\frac{-1991 - a}{2}$ and so the least value is $(\frac{-1991 - a}{2} + 1991)( \frac{-1991 - a}{2} + a) = \frac{1991 - a}{2} \times \frac{-1991 + a}{2} = -\frac{(a-1991)^2}{4}$ . Let this value equal $-1$ and we have the equality $(a-1991)^2 = 4$ . By expansion and Vieta's formula, the sum of these two roots are $2\times1991 = \boxed{3982}$

$(x+1991)(x+a)+1=n^2$

So

$(x+1991)(x+a)=n^2-1=(n+1)(n-1)$

That'll give 2 answers to $a$ . You can check by completing the square.

$(x+1991)(x+a) = -1$

Now, there are only two ways to express -1 as a product, i.e. $-1 \times 1, 1 \times -1$

Case 1:

$x+1991= -1 \\ x+a = 1$

Subtracting the latter equation from the former, we get $1991 - a = -2 \Rightarrow a = 1993$

Case 2:

$x+1991 = 1 \\ x+a = -1$

Again, subtracting the latter equation from the former, $1991-a = 2 \Rightarrow a =1989$

Sum of solutions = $1989+ 1993 = 3982 _{\square}$