How are both integers

Let $k$ be an integer such that $\frac{1-x}{1+x}=k$ .Then we have $x=\frac{1-k}{1+k}$ .Substitute this to the first fraction,we have the fraction $\frac{-k^2-k+1}{2k^2-6k-1}$ .For this to be an integer,we must have $\begin{aligned} |2k^2-6k-1|\leq |-k^2-k+1| \\ \Rightarrow (k^2-7k)(3k^2-5k-2)\leq 0 \\ \Rightarrow k(k-7)(3k+1)(k-2)\leq 0 \end{aligned}$ which have solutions $-\frac{1}{3}\leq k\leq 0$ and $2\leq k\leq 7$ .Substituting the possibilities of $k$ ,we have $k=0,2,3,7$ satisfying the requirement,with respective values of $x=1,-\frac{1}{3},-\frac{1}{2},-\frac{3}{4}$ .Hence,the sum of the absolute values of $x$ is $1+\frac{1}{3}+\frac{1}{2}+\frac{3}{4}=\frac{31}{12}$ ,which yields the answer $31+12=43$

Starting with the easier fraction, we know that $\frac{1-x}{1+x}=N$ for some integer $N$ . Solving for $x$ gives $x=\frac{2}{N+1}-1$ . Since the first fraction must also be integral for the same value of $x$ , it follows that $\frac{(\frac{2}{N+1}-1)^2+4(\frac{2}{N+1}-1)-1}{7(\frac{2}{N+1}-1)^2-6(\frac{2}{N+1}-1)-5}\in\mathbb{Z}$ . Basic algebraic manipulation yields $\frac{-N^2-N+1}{2N^2-6N-1}\in\mathbb{Z}$ for some integer $N$ .

Now some critical intervals of this function will be tested. Let $f(x)=\frac{-x^2-x+1}{2x^2-6x-1}$ To find the asymptote of the function, note that $\lim_{x\to\pm\infty} f(x)=-1/2$ . From $(-\infty<x\le-2$ , $-1<y<0$ , so no solutions exist on this interval. From $8\le{x}<\infty$ , $-1<y<0$ , so no solutions exist on this interval. From $-1\le{x}\le{7}$ , each individual number will be tested. The results of this are:

$\begin{array}{lcr} x & |& f(x )\\ -1 &|& \frac{1}{7}\\ 0 &| &-1\\ 1 &| &\frac{1}{5}\\ 2 &| &1\\ 3 &| &11 \\ 4 &| &-\frac{19}{7}\\ 5& |& -\frac{29}{19}\\ 6 &|&-\frac{41}{35}\\ 7 &|& -1 \end{array}$

As you can see, integer solutions exist only for $N=0,2,3,7$ . From this, we can see that the values of $x$ are $\frac{2}{N+1}-1=\frac{2}{0+1}-1,\frac{2}{2+1}-1,\frac{2}{3+1}-1,\frac{2}{7+1}-1=1,-\frac{1}{3},-\frac{1}{2},-\frac{3}{4}$ . Summing the absolute vaules of these yields $1+\frac{1}{3}+\frac{1}{2}+\frac{3}{4}=\frac{31}{12}$ . $31+12=43$ .

Let

$\frac {x^2+4x-1}{7x^2-6x-5}$ ---------eqn A

$\frac {1-x}{1+x}$ --------eqn B

Analyzing eqn B,

Let $\frac {1-x}{1+x} = p$
if p $\in Z$ then
$|1+x|\leq|1-x|$ for some x, $x \neq 1$
when $x>1$
$|1+x|>|1-x|$
$\Rightarrow x \leq 1$ ( Upper bound ) if p must be an integer

Analyzing eqn A,

$\frac {x^2+4x-1}{7x^2-6x-5} = 0$ has roots at ( $-\sqrt{5}-2,\sqrt{5}-2$ )
for all $x<-\sqrt{5}-2$ , $0 < eqn(A) < 1$
proof:
At $x<-\sqrt{5}-2$ ,both $x^2+4x-1$ and $7x^2-6x-5$ are positive
for eqn A to be > 1
$|x^2+4x-1|>|7x^2-6x-5|$
as both are positive
$x^2+4x-1>7x^2-6x-5$
$6x^2-10x-4 < 0$
$-\frac{1}{3} < x < 2$ which does not tally with $x<-\sqrt{5} -2$

Hence we have a lower bound for x at $\sqrt{5} - 2$ since eqn A never evaluates to an integer at $x<-\sqrt{5} - 2$

Range : $-\sqrt{5}-2 \leq x \leq 1$

However, in this range eqn A and eqn B both have singularities and both can take infinite integer results (not necessarily at the same value of x)

To reduce the range, we know that eqn A = 0 at $\sqrt{5}-2$ . The next integer result that eqn A can take within this range is $-1$

$\frac {x^2+4x-1}{7x^2-6x-5} = -1$
$x=-\frac{3}{4},x=1$

Range : $-\frac{3}{4}\leq x \leq 1$

In this range, eqn B has a continous curve

At $x=1, eqn B=0$
At $x=-\frac{3}{4}, eqn B=7$

 B    |    x   |     A    
 0    |    1   |    -1   
 1    |    0   |    0.2 
 2    |   -1/3 |    1  
 3    |   -1/2 |    11  
 4    |   -3/5 |    -2.714 
 5    |   -2/3 |    -1.526    
 6    |   -5/7 |    -1.171  
 7    |   -3/4 |    -1

$1+\frac{1}{3}+\frac{1}{2}+\frac{3}{4} = \frac{31}{12}$
$\frac{a}{b} = \frac{31}{12}$
$a+b=31+12= 43$

From $\frac{1-x}{1+x}=n$ obtain $x=\frac{1-n}{1+n}$ where $n$ is integer. Substitute this value of x in the first fraction to obtain that $\frac{n^2+n-1}{2n^2-6n-1}$ must be integer. $|n^2+n-1|\ge |2n^2-6n-1|$ is equivalent to $(n^2+n-1)^2-(2n^2-6n-1)^2 \ge 0$ which has solutions (for integer $n)$ the numbers $n=0,n=2,\cdots,7$ . Checking this numbers we see that the fraction is integer only when $n=0,n=2,n=3,n=7$ . Now calculate $x=\frac{1-n}{1+n}$ and add their absolute values to get $1+\frac{1}{3}+\frac{1}{2}+\frac{3}{4}=\frac{31}{12}$ .

Let $k(1+x)=1-x$ . Then, solving, we have $k+kx=1-x$ , $(k+1)x=1-k$ , $x=\frac{1-k}{k+1}$ . Letting $j=k+1$ gives the simpler $x=\frac{2-j}{j}$ . Let $x(j)=\frac{2-j}{j}$ . Then $x'(j)=\frac{(2-j)'*j-(j)'*(2-j)}{j^2}=\frac{(-1)*j-(1)*(2-j)}{j^2}=\frac{-j-2+j}{j^2}=-\frac{2}{j^2}$ . Therefore $x'$ is always negative, and x decreases as j increases. Splitting into two cases, we have when $j>0$ that $x \leq \frac{2-1}{1}=1$ , and when $j<0$ , $x \geq \frac{2-(-1)}{-1}=\frac{3}{-1}=-3$ . Also, $\frac{2-j}{j}+1=\frac{2-j+j}{j}=\frac{2}{j}$ , which is $<0$ when $j<0$ (meaning that $\frac{2-j}{j}+1<0$ and $\frac{2-j}{j}<-1$ , so $x<-1$ ), and is $>0$ when $j>0$ (meaning, similarly, that $x>-1$ ). Combining all together (and neglecting $x \neq 1$ ), we have the nice simple inequality $-3 \leq x \leq 1$ . Now, we know by the Mean Value Theorem that if a function is not monotone on an interval, either a) its derivative is 0 somewhere in the interval, or b) it is discontinuous somewhere in the interval. We will check both for the fraction $\frac{x^2+4x-1}{7x^2-6x-5}$ . a) Let the fraction be $f(x)$ to simplify notation. Then $f'(x)=\frac{(x^2+4x-1)'*(7x^2-6x-5)-(7x^2-6x-5)'*(x^2+4x-1)}{(7x^2-6x-5)^2}$ $=\frac{(2x+4)(7x^2-6x-5)-(14x-6)(x^2+4x-1)}{(7x^2-6x-5)^2}$ , which we shall check for zeros. Setting it equal to 0 and multiplying away the denominator gives us $(2x+4)(7x^2-6x-5)-(14x-6)(x^2+4x-1)=0$ . Expanding the LHS gives $14x^3+16x^2-34x-20-(14x^3+50x^2-38x+6)$ $=-34x^2+4x-26=0$ . The discriminant of this quadratic is $16-4*26*34 \ll 0$ , so this equation has no roots. b) A rational function is discontinuous only when its denominator is 0, so we solve $7x^2-6x-5=0$ to yield the solutions $\frac{3 \pm 2\sqrt{11}}{7}$ , which are approximately $-0.52$ and $1.38$ . Therefore $f(x)$ is monotone in two distinct subintervals of [-3,1]; [-3, -0.52...] and [-0.52..., 1]. Now we just plug in values for j and see where x and f(x) are! We only care about places where integer values are possible. We already know that the first interval applies for $j<0$ (since all x values there are $<-1<-0.52...$ . Checking $j>0$ , we find that the only values in the second interval are where $j=1,2,3,4$ (In this case, we find that $x=1,0,-\frac{1}{3}, -\frac{1}{2}$ ). In the first interval, when $j<0$ we have $f(-3)=-\frac{1}{19} \geq f(x) > f(-1)=-\frac{1}{2}$ , an interval with no integer values. When $j>4$ , the values of x are between $x(5)=-\frac{3}{5}$ and $-1$ , so $f(-\frac{3}{5})=-\frac{19}{7} \leq f(x) < -\frac{1}{2}=f(-1)$ . This interval has two integer values in it, and although -2 is not reachable (I won't show you, but there are two consecutive values straddling -2), -1 certainly is when $j=8$ and $x=-\frac{3}{4}$ . Therefore $-\frac{3}{4}$ is a potential value of x. Quickly checking the four values of x corresponding to $j=1,2,3,4$ , we find surprisingly that 3 of them work: $f(1)=-1$ , $f(-\frac{1}{3})=1$ , and $f(-\frac{1}{2})=11$ . We have 4 values of x, and have rigorously shown that these are the only 4 values. Therefore the sum of their absolute values is $\frac{3}{4}+1+\frac{1}{3}+\frac{1}{2}=\frac{31}{12}$ , and our answer is $31+12=43$ .

This solution, though long, relies on one simple, intuitive fact: Most functions are monotone most of the time!

First, we take $\frac{1-x}{1+x}=b$ (b is an integer) $\rightarrow$ $\frac{1-b}{1+b}=x$ . We then replace x in the first factor and receive another factor: $\frac{b^2+b-1}{1+6b-2b^2}=a$ ( a is an integer) $\rightarrow\ (2a+1)b^2+(1-6a)b-1-a=0$ . This equation has two roots . Now we find a for the first root $b=\frac{6a-1+\sqrt{44a^2+5}}{2(2a+1)}$ is integer. Note that $\frac{6a-1+\sqrt{44a^2+5}}{2(2a+1)}=\frac{3}{2}+\frac{\sqrt{44a^2+5}-4}{2(2a+1)}$ . We can infer that 2a+1 must divides $\sqrt{44a^2+5}$ . If $|a|>11$ we have the inequality : $4(2a+1)>\sqrt{44a^2+5}-4>3(2a+1)$ , a contradiction so $|a|\leq11$ . Just calculate and we will have $a=\pm1$ and $a=11$ that give $x=\frac{-1}{3},\frac{-1}{2},1,\frac{-3}{4}$ . The another roots of b is just the same but it has no integer a that satisfy b is integer.

Note : m^n denotes m to the power of n

Let f(x) = ( x^2 + 4 * x - 1 ) / ( 7 * x^2 - 6 * x - 5 )

and g(x) = ( 1 - x / 1 + x )

Claim : g(x) is not a rational number for any irrational value of x.

Proof : Let there exists an irrational no. x0 such that g(x0) is equal to an

        integer say r. Then 

            r = ( 1 - x0 ) / ( 1 + x0)

        which implies 

       x0 = ( 1 - r ) / ( 1 + r )

which is rational since r is rational.

This implies x0 is rational which is a contradiction to our assumption.

Hence, g(x) cannot be rational and therefore an integer for any irrational value of x.

Now, g(1) = 0 and f(1) = -1 , so x = 1 is a solution.

Now, modulus of any non-zero integer is greater than 1.

So, if g(x) is an integer, then

       | 1 - x | ≥ | 1 + x  |

From which we get either x < -1 or -1 < x ≤ 0

Let x = m / n where n is a natural number and m is an integer such that

gcd ( m , n ) = 1

Also since | x | < 1 it implies that | m | < n

Then g(x) = ( n - m ) / ( n + m ) = 1 - ( 2 * m / m + n )

This will be an integer in the following cases

i) m + n = 1

but n ≥ 1 implies m ≤ 0

So putting x = ( 1 - n ) / n we get g ( x ) = 2 * n - 1 for natural no. n

ii) m + n = -1 but n ≥ | m | implies m + n ≥ 0, so this case is invalid

iii) m + n = 2 but n ≥ 1 implies m ≤ 1

So putting x = ( 2 - n ) / n we get g (x ) = n - 1 for natural no. n

iv) m + n = -2 which is again invalid since m + n ≥ 0

v) m + n = 2 * m i.e. m = n i.e. x = 1

vi) m + n = -2 * m i.e. n = -3 * m
or x = - 1 / 3 and g( x ) = 2

but this solution is included in the solution of case ( iii ) i.e. m + n = 2

Now, of the above values of x we have to find those which make f(x) an integer.

i) x = 1 for which g ( x ) = 0 and f ( x ) = -1

ii) x = ( 1 - n ) / n for natural n or putting 1 - n = -k for whole no. k i.e. x = -k / ( 1 + k ) we get

f ( x ) = - ( 4 * k ^2 + 6 * k + 1 ) / ( 8 * k^2 - 4 * k - 5 )

For f(x) to be an integer | f ( x ) | ≥ 1 ( since f ( x ) ≠ 0 for any x )

  |- ( 4 * k ^2 + 6 * k + 1 ) | ≥  | ( 8 * k^2 - 4 * k - 5 ) |

But 4 * k ^2 + 6 * k + 1 > 0 for all natural k and

    8 * k^2 - 4 * k - 5 > 0       for all natural k > 1

So, for k = 1 we get x = -1 / 2 and f (x) = 11

and for k > 1 we get

9 * k^2 - 5 * k - 14 ≤ 0

This means 1 < k ≤ 3 but for k = 2 ,

f(x) is not an integer and for k = 3

x = -3 / 4 and f(x) = -1

iii) x = ( 2 - n ) / n for natural n or putting 2 - n = -k for integer k

i.e. x = -k / ( 2 + k ) we get

f ( x ) = - 4 * ( k ^2 + 3 * k + 1 ) / ( 13 * k^2 + 7 * k - 10 )

For f(x) to be an integer | f ( x ) | ≥ 1 which means

| - 4 * ( k ^2 + 3 * k + 1 ) | ≥ | ( 13 * k^2 + 7 * k - 10 ) |

Solving similarly as above we get two values of k which are

k = -1 i.e. x = 1 for which f(x) = -1 and

k = 1 i.e. x = -1 / 3 for which f(x) = 1

Hence , all the values of x for which f(x) and g(x) are integers are

x = 1 ,- 1 / 2 , - 1 / 3, - 3 / 4

The sum of absolute values of the above values of x = 31 / 12.

Since gcd ( 31 , 12 ) = 1, a = 31 and b =12.

Hence a + b = 43 which is the required answer.

$\frac{1-x}{1+x} = \frac{2}{1+x}-1 \in \mathbb{Z} \Rightarrow \frac{2}{1+x} = k \in \mathbb{Z} \Rightarrow x=\frac{2}{k}-1$

Substituting this into the first fraction, $\frac{x^2+4x-1}{7x^2-6x-5} = \frac{(\frac{2}{k}-1)^2+4(\frac{2}{k}-1)-1}{7(\frac{2}{k}-1)^2-6(\frac{2}{k}-1)-5} \times \frac{k^2}{k^2}$ = \frac{-4k^2+4k+4}{8k^2-40k+28} = \frac{-k^2+k+1}{2k^2-10k+7})

My solution is not strictly under Mathematics I first converted these two fractions in one fraction of k(which is integer) which yields integer

Conversion of fraction

Then i run loop of k from 0 to 100 to gain values of k and x

Let $\frac{1-x}{1+x} = a$ and $\frac{x^2+4x-1}{7x^2-6x-5} = b$ with a and b both being integers. Then $1-x = a + ax\Rightarrow x = \frac{1-a}{1+a}$

Plug $x = \frac{1-a}{1+a}$ into $\frac{x^2+4x-1}{7x^2-6x-5}= b$ , we get $\frac{1-a-a^a}{2a^a-6a-1} = b$

$1-a-a^2 = (2b)a^2+(-6b)a+(-b) \Rightarrow 0 = (2b+1)a^2 + (1-6b) a +(-b-1)$

Now, use the quadratic formula. Since $a$ is an integer, the discriminant must be a perfect square:

$(1-6b)^2 -4(2b+1)(-b-1) \Rightarrow 44b^2 + 5 = 0$

$a=\frac{(6b-1)\pm\sqrt{44b^2 + 5}}{4b+1}$

After trying a few values of b, we find that for $b=\pm1$ the discriminant is 7, and for $b = \pm11$ the discriminant is 73.

Now, we plug b back into the equation $a=\frac{(6b-1)\pm\sqrt{44b^2 + 5}}{4b+1}$ :

$b=1$ $a = 2,\frac{-1}{3}$

$b=-1$ $a=0,7$

$b=11$ $a=3,\frac{-4}{23}$

$b=-11$ $a=\frac{-1}{7},\frac{10}{3}$

Taking only the integer solutions, $a=0,2,3,7$ . Plug this back into $x = \frac{1-a}{1+a}$ .

$x = \frac{-1}{2},\frac{-1}{3},\frac{-3}{4},1$

Adding the absolute value of each x gives $\frac{31}{12}$ , so the answer is 43 .

Analisemos a segunda fracção, uma vez que é menos 'complexa'!!

• Se fizermos $x = 0$ , teremos um inteiro, no entanto, não na primeira fracção;
• $x = - 1$ , anula o denominador, portanto impossível;
• $x \leq - 2$ , não resultará num $\mathbb{Z}$ ;
• Ao substituir "x" por 1, isto é, $x = 1$ ambas as fracções serão inteiras, por essa razão substituir aquele valor na 1ª fracção obtemos $- 1$ , logo,

$\frac{x^2 + 4x - 1}{7x^2 - 6x - 5} = - 1 \\\\ 4x^2 - x - 3 = 0 \\\\ \left ( x - 1 \right )\left ( x + \frac{3}{4} \right ) = 0$

Aplicando raciocínio análogo, obtemos: 1, - 3/4, - 1/2 e - 1/3.

Portanto,

$\left | 1 \right | + \left | - \frac{3}{4} \right | + \left | - \frac{1}{2} \right | + \left | - \frac{1}{3} \right | = \\\\ 1 + \frac{3}{4} + \frac{1}{2} + \frac{1}{3} = \\\\ \frac{31}{12}$

Daí,

$a + b = 31 + 12 \\\\ \boxed{\boxed{a + b = 43}}$