What integer makes $(x-a)(x-10) +1$ factorable as $(x+b)(x+ c)$?

[kaliprasad]

find integer a such that $(x-a)(x-10) +1$ can be factored as $(x+b)(x+ c)$ where b and c are integers

 

[Prove It]

find integer a such that $(x-a)(x-10) +1$ can be factored as $(x+b)(x+ c)$ where b and c are integers

$\displaystyle \begin{align*} \left( x - a \right) \left( x - 10 \right) + 1 &= \left( x + b \right) \left( x + c \right) \\ x^2 - 10x - a\,x + 10a + 1 &= x^2 + c\,x + b\,x + b\,c \\ x^2 - \left( 10 + a \right) x + 10a + 1 &= x^2 + \left( b + c \right) x + b\,c \end{align*}$

So equating the coefficients gives $\displaystyle \begin{align*} - \left( 10 + a \right) = b + c \end{align*}$ and $\displaystyle \begin{align*} 10a + 1 = b\,c \end{align*}$. See what you can do from here.

 

[anemone]

$\displaystyle \begin{align*} \left( x - a \right) \left( x - 10 \right) + 1 &= \left( x + b \right) \left( x + c \right) \\ x^2 - 10x - a\,x + 10a + 1 &= x^2 + c\,x + b\,x + b\,c \\ x^2 - \left( 10 + a \right) x + 10a + 1 &= x^2 + \left( b + c \right) x + b\,c \end{align*}$

So equating the coefficients gives $\displaystyle \begin{align*} - \left( 10 + a \right) = b + c \end{align*}$ and $\displaystyle \begin{align*} 10a + 1 = b\,c \end{align*}$. See what you can do from here.

Hi Prove It,(Smile)

This problem is meant to be a challenge and thus, it deserves a fully well explained solution. But I understand that you have been posting quite much help at this site recently that it might be just the case that you didn't aware this problem comes from the Challenge Problems sub-forum.:)

 

[Prove It]

Hi Prove It,(Smile)

This problem is meant to be a challenge and thus, it deserves a fully well explained solution. But I understand that you have been posting quite much help at this site recently that it might be just the case that you didn't aware this problem comes from the Challenge Problems sub-forum.:)

Yes I should look at what forum these questions are in hahaha.

 

[anemone]

find integer a such that $(x-a)(x-10) +1$ can be factored as $(x+b)(x+ c)$ where b and c are integers

Hi kaliprasad,

Thanks for posting another great math challenge at MHB, I deeply appreciate that.

For this challenge, I was wondering if there are (finitely) many values for $a$ that meet the condition of the problem? Also, if $b$ and $c$ meant to be distinct integer? If they are not, then I can tell $a=8$ is a valid solution though.:)

 

[kaliprasad]

Hi kaliprasad,

Thanks for posting another great math challenge at MHB, I deeply appreciate that.

For this challenge, I was wondering if there are (finitely) many values for $a$ that meet the condition of the problem? Also, if $b$ and $c$ meant to be distinct integer? If they are not, then I can tell $a=8$ is a valid solution though.:)

Not a solution but resolving query. you should provide the method

Spoiler

1) there are more than 1
2) b and c may be same
3) finite number of solutions

 

[Saitama]

Spoiler

$$(x-a)(x-10)+1=x^2-(10+a)x+10a+1$$
Since we need to resolve the above into two linear factors, the discriminant must be greater than or equal to zero i.e
$$(10+a)^2-4\cdot (10a+1) \geq 0 \Rightarrow a^2-20a+96 \geq 0$$
$$\Rightarrow (a-12)(a-8) \geq 0$$
Also, the discriminant must be a perfect square i.e
$$(a-12)(a-8)=m^2$$
where $m$ is some integer. Substitute $a-10=t$.
$$\Rightarrow t^2=2^2+m^2$$
Clearly, $m=0$ i.e $\boxed{a=8,12}$.

 

[kaliprasad]

Spoiler

$$(x-a)(x-10)+1=x^2-(10+a)x+10a+1$$
Since we need to resolve the above into two linear factors, the discriminant must be greater than or equal to zero i.e
$$(10+a)^2-4\cdot (10a+1) \geq 0 \Rightarrow a^2-20a+96 \geq 0$$
$$\Rightarrow (a-12)(a-8) \geq 0$$
Also, the discriminant must be a perfect square i.e
$$(a-12)(a-8)=m^2$$
where $m$ is some integer. Substitute $a-10=t$.
$$\Rightarrow t^2=2^2+m^2$$
Clearly, $m=0$ i.e $\boxed{a=8,12}$.

above is a good ans and completely different from solution below

Spoiler

as $(x-a)(x-10)+ 1 = (x+b)(x+c)$
Putting $x = -b$ we get
$(-a-b)(-b-10) + 1=0$ or
$(a+b)(b+10) = -1$
This implies $(a+b) = 1$ and $b+ 10 = - 1$ or $b = -11\, a = 12$
Or $a+b = -1$ and $b + 10 = 1$ => $b = -9\, a = 8$
hence a = 8 or 12

because of symmetry nature of RHS we need not put x = c