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BdMO Online Forum • View topic - Find $$(x,y)$$

## Find $$(x,y)$$

For students of class 9-10 (age 14-16)

### Find $$(x,y)$$

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jayon

Posts: 20
Joined: Thu Dec 08, 2016 12:10 am

### Re: Find $$(x,y)$$

The only solution is $(x,y) = (3,11)$

Solution:
First, I tried to divide $x^5 - 1$ by $x-1$. It became,

$y^2 = \frac{x^5-1}{x-1}$
$\Leftrightarrow y^2 = x^4 + x^3 + x^2 + x + 1$

Then, I think to put an inequality and to find an easy equation so that I can find the value of $x$.

Notice that, $(x^2 + \frac{x}{2} )^2 = x^4 + x^3 + \frac{x^2}{4}$
and, $( x^2 + \frac{x}{2} + 1 )^2 = x^4 + x^3 + \frac{9x^2}{4} + x + 1$

Now, $x^4 + x^3 + \frac{x^2}{4} < x^4 + x^3 + x^2 + x + 1 < x^4 + x^3 + \frac{9x^2}{4} + x + 1$
$\Rightarrow (x^2 + \frac{x}{2} )^2 < y^2 < ( x^2 + \frac{x}{2} + 1 )^2$
$\Rightarrow x^2 + \frac{x}{2} < y < x^2 + \frac{x}{2} + 1$

If $x = 2k, k\in\mathbb{N}$, then it'll be,

$4k^2+k<y<4k^2+k+1 \Longrightarrow y\notin\mathbb{N}$ $[\because (x,y) \in\mathbb{N} ]$

So, there is no solution if $y$ is even number.

If $x=2k+1,k\in\mathbb{N}$, then it'll be,

$4k^2+5k+\frac{1}{2} + 1<y<4k^2+5k+1+\frac{1}{2}+1$
$\Rightarrow y = 4k^2+5k+2 = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2}$
$\Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2$
$\Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4}$
$\Rightarrow x^2 - 2x - 3 = 0$
$\Rightarrow (x - 3)(x + 1) = 0$

So, $x = 3 , y = 11$

I'll really appreciate if anyone check the solution to tell me if it's correct or not.
super boy

Posts: 5
Joined: Tue Dec 20, 2016 1:11 pm

### Re: Find $$(x,y)$$

super boy wrote:The only solution is $(x,y) = (3,11)$

Solution:
First, I tried to divide $x^5 - 1$ by $x-1$. It became,

$y^2 = \frac{x^5-1}{x-1}$
$\Leftrightarrow y^2 = x^4 + x^3 + x^2 + x + 1$

Then, I think to put an inequality and to find an easy equation so that I can find the value of $x$.

Notice that, $(x^2 + \frac{x}{2} )^2 = x^4 + x^3 + \frac{x^2}{4}$
and, $( x^2 + \frac{x}{2} + 1 )^2 = x^4 + x^3 + \frac{9x^2}{4} + x + 1$

Now, $x^4 + x^3 + \frac{x^2}{4} < x^4 + x^3 + x^2 + x + 1 < x^4 + x^3 + \frac{9x^2}{4} + x + 1$
$\Rightarrow (x^2 + \frac{x}{2} )^2 < y^2 < ( x^2 + \frac{x}{2} + 1 )^2$
$\Rightarrow x^2 + \frac{x}{2} < y < x^2 + \frac{x}{2} + 1$

If $x = 2k, k\in\mathbb{N}$, then it'll be,

$4k^2+k<y<4k^2+k+1 \Longrightarrow y\notin\mathbb{N}$ $[\because (x,y) \in\mathbb{N} ]$

So, there is no solution if $y$ is even number.

If $x=2k+1,k\in\mathbb{N}$, then it'll be,

$4k^2+5k+\frac{1}{2} + 1<y<4k^2+5k+1+\frac{1}{2}+1$
$\Rightarrow y = 4k^2+5k+2$ = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2}  \Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2  \Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4}  \Rightarrow x^2 - 2x - 3 = 0  \Rightarrow (x - 3)(x + 1) = 0 $So,$ x = 3 , y = 11 $I'll really appreciate if anyone check the solution to tell me if it's correct or not. HEI.. WHAT YOU DID WRITE IN THE LAST PART.. PLZ MAKE IT CLEAR TO UNDERSTAND... DIDN'T YOU USE LATEX fonts..IF U DID THEN TRY TO USE IT PROPERLY DUDE... jayon Posts: 20 Joined: Thu Dec 08, 2016 12:10 am ### Re: Find $$(x,y)$$ He just forgotten to give dollar . For him I am giving that..... The last part would be$ y=4k^2+5k+2 = 4k^2+5k+1+\frac{1}{2}+\frac{1}{2}
\Rightarrow y^2 = (x^2 + \frac{x}{2} + \frac{1}{2} )^2
\Rightarrow x^4 + x^3 + x^2 + x + 1 = x^4 + x^3 + x^2 + \frac{x^2}{4} + \frac{x}{2} + \frac{1}{4}
\Rightarrow x^2 - 2x - 3 = 0
\Rightarrow (x - 3)(x + 1) = 0 So x = 3 , y = 11 \$
soyeb pervez jim

Posts: 2
Joined: Sat Jan 28, 2017 11:06 pm