If If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be which of the given choices?

Question

If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be

(A) 50 (B) 37 (C) 19 (D) 61

My approach: I first took roots $\alpha$, $\beta$ and then applied sum and product of roots according to Vieta's formula. But I don't think we can get any value with that. Any help will be much appreciation.

Also can we actually find the range of $a^2+b^2$.

Answer 1

Now this equation is having positive integer roots so the sum as well as product of the roots should be positive integer. $\alpha + \beta = -a \gt 0$ and $\alpha \beta = b+1 \gt 0$. Which gives us the two conditions: $$a \lt 0$$ and $$b \gt -1$$ Now since the equation is having positive integers as roots so sum of roots is a positive integer implying $a$ is an integer and the product is also an integer implying $b$ is also an integer.

Now for the equation $a^2+b^2$ we can only have choices where $a$ is a negative integer and $b$ is an integer greater than $-1$.

For the first choice $a^2+b^2=50$ you will get the values of $(a,b)$ as $(-1,7)$$(-5,5)$ and $(-7,1)$ out of which only $(-5,5)$ satisfies.

For the second choice $a^2+b^2=37$, you get the values of $(a,b)$ as $(-1,6)$ and $(-6,1)$ but none of them will satisfy since the discriminant is not a perfect square.

For the third choice there is no such $(a,b)$ and for the fourth option the values of $(a,b)$ are $(-5,6)$ and $(-6,5)$ and none of them actually have a perfect square discriminant.

So option A is correct. Hope this helps ….

Answer 2

We need that $a$ and $b$ will be naturals and there is a natural $n$ for which $$a^2-4(b+1)=n^2.$$ Thus, $$a^2+b^2=n^2+b^2+4b+4=n^2+(b+2)^2.$$ Can you end it now?