Are quadratics that easy?

The given quadratic equation can be rewritten as $(a^{2}+b^{2})x^{2}-(4ab+1)x+(a^{2}+b^{2}) = 0$ Since one of the roots is an integer, the two roots are real numbers.

So the discriminant of the quadratic is non-negative. $(4ab+1)^{2}-4(a^{2}+b^{2})^{2} \ge 0$ $(1+2(a+b)^{2})(1-2(a-b)^{2}) \ge 0$ Since the first factor $(1+2(a+b)^{2})$ is always positive, the second factor must be non-negative. $1-2(a-b)^{2} \ge 0$ $(a-b)^{2} \le \frac{1}{2}$ Since $a$ and $b$ are non-zero integers, the value of $(a-b)$ must be zero.

Thus, $a = b$

Then the given quadratic can be written as $2a^{2}x^{2}-(4a^{2}+1)x+2a^{2} = 0$

Suppose the roots of the quadratic are $x_{1}$ and $x_{2}$ ; where $x_{1}$ is the integer root.

By Vieta's formula, $x_{1}x_{2} = 1$ $x_{1}+x_{2} = 2+\frac{1}{2a^{2}}$

Note that $x_{1} \ne 0$ and $x_{1} \ne 1$ . So $x_{1} \ge 2$ $x_{2} = \frac{1}{x_{1}} > 0$ $2 \le x_{1} < x_{1}+x_{2}$ As $a$ is a non-zero integer, thus $a^{2} \ge 1$ $x_{1}+x_{2} = 2+\frac{1}{2a^{2}} \le 2.5$ $2 \le x_{1} < x_{1}+x_{2} \le 2.5$ As $x_{1}$ is an integer, so $x_{1} = 2$ and $x_{2} = \frac{1}{2}$

The sum of the roots is $2+\frac{1}{2} = \frac{5}{2} = \frac{m}{n}$ $m+n = 5+2 = 7.\square$

The quadratic equation becomes:

$(a^{2}+b^{2})x^{2}-(4ab+1)x+(a^{2}+b^{2})=0$

Since one root is integer, hence Discriminant,

$D=B^{2}-4AC ≥0$

$(1+4ab)^{2}-(2(a^{2}+b^{2}))^{2}≥0$

$(1+2(a+b)^{2})(1-2(a-b)^{2})≥0$

Since first bracket is always +ve, the second bracket needs to be ≥0

$1-2(a-b)^{2}≥0$

$(a-b)^{2}≤1/2$

Since a & b are non-zero integers, hence,

$(a-b)^{2}=0$ or $a=b$

Now, let roots of given quadratic be $x_{1}$ (integer) and $x_{2}$

Sum of roots = -B/A and Product of roots = C/A

$x_{1}+x_{2}= (4a.a+1)/(a^{2}+a^{2})=2+1/2a^{2}$

$x_{1}x_{2}=1$ or $x_{2}=1/ x_{1}$

$x_{1}+1/x_{1}=2+1/2a^{2}$

Since a is non-zero integer, $a^{2} ≥1$

$2+1/2a^{2}≤2+1/2=2.5$

$2 ≤ x_{1}+1/x_{1}≤2.5$

Hence,

$x_{1}=2$ and $x_{2}=1/2$

$x_{1}+x_{2}=5/2=m/n$

$m+n=7$

Let me attempt a proof with inequalities.

Lemma. For all reals $m, n, m^{2}+n^{2} \geq 0.5(m+n)^{2}$ This is a result of Cauchy Schwarz inequality.

Let $m=ax-b, n=bx-a$ so $x \geq 0.5(a+b)^{2}(x-1)^{2}$ and thus $(a+b)^{2} \leq \frac {2x}{(x-1)^{2}}$ . We let $\frac {2x}{(x-1)^{2}}$ be $y$ .

Consider $y$ . If $x> 3, y <1$ so $a+b<1 \implies a=-b$ . Substituting, $2a^{2}(x+1)^{2}=x \geq 2 (x+1)^{2}$ , contradiction. Hence $x=1, 2, 3$ . (x is positive) But 3 cannot be written as the sum of two perfect squares, so we consider 1 and 2.

$x=1 \implies 2 (a-b)^{2}=1$ which is a contradiction.

$x=2 \implies (2a-b)^{2}+(2b-a)^{2}=2$ . This gives $(2a-b)^{2}=(2b-a)^{2}=1$ . WLOG let $a \geq b$ . This gives the solutions $(a, b)=(1, 1), (\frac {1}{3}, -\frac {1}{3}), (-1, -1)$ . Note that $(1, 1), (1, -1)$ give the same x. So we just need to consider $(1, 1)$

The equation is now $2 (x-1)^{2}=x$ which has two solutions 2 and 0.5. Hence their sum is $\frac {5}{2}$ and the answer is 7.

On rearranging $(ax-b)^{2} + (bx-a)^{2} = x$ we get $(a^{2}+b^{2})x^{2} - (4ab+1)x + (a^{2}+b^{2}) = 0$

The roots of the above equation are: $((4ab+1) +/- \sqrt((4ab+1)^{2} - 4(a^{2}+b^{2})^2)) / (2(a^{2}+b^{2})$

It's given that x has an integer solution. So we can eliminate the complex solutions. That means to say, $(4ab+1)^{2} > 4((a^{2} + b^{2}))^{2}$ i.e. $(4ab+1)^{2} > (2(a^{2} + b^{2}))^{2}$

It's given that a and b are non-zero integers. By giving various values for a and b in the above inequality, we see that the inequality is satisfied only when a = b.

With a=1, b=1, we get x = 2 or 1/2.

Sum of roots of x is 2+(1/2) = 5/2.

5 and 2 are co-prime (which is one of the conditions for the solution: m/n, m and n are co-prime)

m+n = 5+2 = 7.

First we notice the left hand side is a non-negative real, so $x\geq0$ . By trial and error we find one possible solution and show that $x\not\in\{0;\:1\}$ : $\begin{aligned} (a_1;\:b_1;\:x_1)&=(1;\:1;\:2),&&&&&x&=0:&a^2+b^2&\neq 0,&&&&&x&=1:&2(a-b)^2&\neq 1\mod 2 \end{aligned}$ Now to the general case. We expand the equation and may divide by $a^2+b^2\geq 2$ , because $a,\:b\neq0$ : $\begin{aligned} 0&=x^2-\frac{4ab +1}{a^2+b^2} x +1=:x^2-2kx+1&&&\Rightarrow &&&&x_{1,2}&=k\pm\sqrt{k^2-1} & (*) \end{aligned}$ To get (at least) one non-negative real solution, we need $k\geq 1$ . Using AM-GM-HM (arithmetic/geometric/harmonic mean inequality) on $k$ : $\begin{aligned} |k|\leq\frac{4|ab|+1}{2(a^2+b^2)}&=\frac{1}{ \frac{1}{2}\left(\frac{|a|}{|b|}+\frac{|b|}{|a|}\right) } + \frac{1}{2(a^2+b^2)}\underset{\text{HM-GM}}{\leq} 1+\frac{1}{2\cdot 2}=\frac{5}{4} &&&\left| a^2+b^2\geq 2\right. \end{aligned}$ With the remaining values for $k$ , try to get an upper estimate for the solutions: $\begin{aligned} k&\in \left[1;\:\frac{5}{4}\right]: &&& |x_{1,2}|&=|k\pm\sqrt{k^2-1}|\leq |k|+\sqrt{k^2-1}\leq\frac{5}{4}+\sqrt{\frac{25}{16}-1}=2 \end{aligned}$ The only possible non-negative integer solutions are $x\in\{0;\:1;\:2\}$ . We already eliminated the first two and showed that $x_1=2$ is indeed a solution to the problem. Vieta's Formula in $(*)$ yields the second root $x_2=\frac{1}{2}$ , so $x_1+x_2=\frac{5}{2}=\frac{m}{n}$ , and the answer is $m+n=\boxed{7}$

x^2 - 2 [(2 ab + 0.5)/(a^2 + b^2)] x + 1 = 0

x = (2 ab + 0.5)/(a^2 + b^2) +/- SQRT{[(2 ab + 0.5)/(a^2 + b^2)]^2 - 1}

Discriminant >= 0

=> [(2 ab + 0.5)/(a^2 + b^2)]^2 - 1 >= 0

=> [(2 ab + 0.5)/(a^2 + b^2) + 1][(2 ab + 0.5)/(a^2 + b^2) - 1] >= 0

=> [(2 ab + 0.5)/(a^2 + b^2) - 1] >= 0

=> 0.5 >= a^2 + b^2 - 2 a b

=> (a - b)^2 <= 0.5

Only integer sets allowed are a - b = 0 where a = b such as (a, a). (a, -a) is also eliminated.

x = (1 + 0.25/ a^2) +/- SQRT[(1 + 0.25/ a^2)^2 - 1]

Discriminant = (0.25/ a^2)(2 + 0.25/ a^2) must always satisfy greater than or equal to zero.

(1 + 0.25/ a^2) - SQRT[(1 + 0.25/ a^2)^2 - 1] is not likely an integer. With a = 1 to 20 and a = -1 to -20 for examples [Where only non-zero (a, a) is wanted],

x = (1 + 0.25/ a^2) + SQRT[(1 + 0.25/ a^2)^2 - 1] or (1 + 0.25/ a^2) - SQRT[(1 + 0.25/ a^2)^2 - 1]

Compute using Excel can be obvious that only (a, a) = (-1, -1) or (1, 1) gives x = 2 or 0.5 to satisfy.

x^2 - (5/2) x + 1 = 0 is the only equation satisfies stated conditions.

Sum of all distinct x = 5/2 and therefore 5 + 2 = 7.