MHB Square Int x: $x(x+1)(x+7)(x+8)$ is Square Integer

[kaliprasad]

Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer

 

[Albert1]

Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer

my solution:

Spoiler

$x(x+1)=(x+7)(x+8)---(1)$
$x(x+7)=(x+1)(x+8)---(2)$
$x(x+8)=(x+1)(x+7)---(3)$
solution of $(1) : x=-4$
solution of $(2) : x=-4$
solution of $(3) : $ no solution
also $x(x+1)(x+7)(x+8)=0---(4)$
solution of $(4) : x=0,-1,-7,-8$
$x(x+1)(x+7)(x+8)=144---(5)$
solution of $(5):x=1,-9$
from above :$x\in$ $\left \{ 1,0,-1,-4,-7,-8,-9 \right \}$

 

[kaliprasad]

my solution:

Spoiler

$x(x+1)(x+7)(x+8)=144---(5)$

Spoiler

How ?

 

[Albert1]

Spoiler

How ?

Spoiler

from (2)
$(x+1)(x+8)=x(x+7)$
$x=-4,(x+1)(x+8)=x(x+7)=-12$
$\therefore x(x+1)(x+7)(x+8)=(-12)^2=144$

 

[kaliprasad]

Spoiler

from (2)
$(x+1)(x+8)=x(x+7)$
$x=-4,(x+1)(x+8)=x(x+7)=-12$
$\therefore x(x+1)(x+7)(x+8)=(-12)^2=144$

I understand .

Spoiler

can u show that there is no other solution

 

[Opalg]

Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer

[sp]Let $y = x+4$. Then $$\begin{aligned} x(x+1)(x+7)(x+8) &= (y-4)(y+4)(y-3)(y+3) \\ &= (y^2-16)(y^2-9) \\ &= y^4 - 25y^2 + 144 \\ &= (y^2-12)^2 - y^2. \end{aligned}$$ If that is a square then either $y=0$ or $y^2$ must be at least as large as the difference between $(y^2-12)^2$ and the previous square $(y^2-13)^2$. But $(y^2-12)^2 - (y^2-13)^2 = 2y^2 - 25.$ So we must have $y^2 \geqslant 2y^2 - 25$, which means that $y^2 \leqslant 25.$ Therefore $y$ lies between $-5$ and $5$, so that $x$ lies between $-9$ and $1$. The valid solutions in that interval are exactly those found by Albert.[/sp]

 

[kaliprasad]

my solution

Spoiler

we see that x = 0, x = -1, x = -7 and x = -8 gives the answer zero so a perfect square
let us look for other values
we have
$x(x+1)(x+7)(x+8)$
= $x(x+8)(x+1)(x+7)$
= $(x^2+8x)(x^2+8x+7)$
$= y(y+7)$ where y is $x^2+8x$
for it to be a perfect square we see that $GCD(y,y+7) = GCD(y,7)$
y cannot be a multiple of 7 because then y and y + 7 are consecutive multiples of 7 and as y is not zero product cannot
be a perfect square.
so y and y + 7 are coprimes and hence perferct squares and both are -ve of perfect squares
taking positive values let $y = n^2$ and $y+7 = m^2$
giving $n^2+7=m^2$
or $m^2-n^2 = 7$
or $(m+n)(m-n) = 7 * 1$ hence $m+n = 7, m-n= 1=>m= 4,n= 3$ giving y = 9 , y + 7 = 16
so $y = 16$
hence $x^2+8x-9=0$ giving $x = 1,=9$
taking -ve values we have $y= - 16 , y + 7 = - 9$
or $x^2+8x+ 16= 0 => x = - 4$
so we have x is one of $-9,-8,-7,-4, -1,0,1$