MHB How to Form a Quadratic Equation with Coefficients in AP and Integer Roots?

[kaliprasad]

form a quadratic equation $ax^2 + bx + c = 0$ such that a,b,c are in AP and it has got integer roots

 

[Nono713]

[sp]

Set $a = 1$ for convenience. Then the quadratic has integer roots if and only if $b^2 - 4c$ is a perfect square. Furthermore we need $a$, $b$ and $c$ to be in arithmetic progression, so we must have $b = 1 + k$ and $c = 1 + 2k$ for some nonzero integer $k$. Thus we need $(1 + k)^2 - 4(1 + 2k) = k^2 - 6k - 3$ to be a perfect square. The first one that works is $k = 7$, since $7^2 - 6 \times 7 - 3 = 4 = 2^2$. Therefore a solution is:
$$x^2 + 8x + 15 = 0$$
Which has integer roots $-5$ and $-3$, and $1$, $8$, $15$ are in arithmetic progression.

TODO: find a systematic way to choose $k$, and prove/disprove there are infinitely many solutions for any $a \ne 0$.​

[/sp]

 

[Opalg]

[sp]

Set $a = 1$ for convenience. Then the quadratic has integer roots if and only if $b^2 - 4c$ is a perfect square. Furthermore we need $a$, $b$ and $c$ to be in arithmetic progression, so we must have $b = 1 + k$ and $c = 1 + 2k$ for some nonzero integer $k$. Thus we need $(1 + k)^2 - 4(1 + 2k) = k^2 - 6k - 3$ to be a perfect square. The first one that works is $k = 7$, since $7^2 - 6 \times 7 - 3 = 4 = 2^2$. Therefore a solution is:
$$x^2 + 8x + 15 = 0$$
Which has integer roots $-5$ and $-3$, and $1$, $8$, $15$ are in arithmetic progression.

TODO: find a systematic way to choose $k$, and prove/disprove there are infinitely many solutions for any $a \ne 0$.​

[/sp]

[sp]In fact, assuming that $a$, $b$ and $c$ are integers there is only one other solution apart from $x^2 + 8x + 15 = 0$ or multiples of it.

If $a$ and $k$ have a common factor $d$ then $d$ will divide $a$, $b$ and $c$, and we can divide the equation $ax^2 + bx + c = 0$ by $d$ to get an equivalent version of the same equation. So we may assume that $a$ and $k$ are coprime. But the sum of the roots (which has to be an integer) is $\frac{-(a+k)}{a} = -1 -\frac ka$. If $a$ and $k$ are coprime that can only be an integer when $a=1$. So the equation becomes $x^2 + (1+k)x + (1+2k)$, with roots $\frac12\bigl( -1-k \pm\sqrt{(k+1)^2 - 4(2k+1)}\bigr).$

The expression under the square root is $k^2 - 6k - 3 = (k-3)^2 - 12$. That has to be the square of some integer $n$. If we write $k-3=m$ then $m$ and $n$ have to satisfy the condition $m^2 - 12 = n^2$. Then $m^2 - n^2 = (m+n)(m-n) = 12.$ The only factorisations of $12$ making $m$ and $n$ both integers are are $12 = (\pm6)(\pm2)$, giving $m = \pm4$ and $n = \pm2.$ If $m=4$ then $k=7$, giving the solution ground by Bacterius. If $m=-4$ then $k=-1$, giving the equation $x^2 + 0x + (-1) = 0$, or $x^2-1=0$, with solutions $x = \pm1.$[/sp]

 

[kaliprasad]

my solution is same as above except the starting point

Spoiler

because the roots are integer say m and n then
$f(x) =l(x+m)(x+n)$

so for the case when gcd( coefficients of $x^2$, coefficients of x, constant) = 1 then coefficient of $x^2 = 1$