Find $$(x,y)$$
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Re: Find $$(x,y)$$
The only solution is $(x,y) = (3,11) $
Solution:
I'll really appreciate if anyone check the solution to tell me if it's correct or not.
Solution:
Re: Find $$(x,y)$$
HEI.. WHAT YOU DID WRITE IN THE LAST PART.. PLZ MAKE IT CLEAR TO UNDERSTAND...super boy wrote:The only solution is $(x,y) = (3,11) $
Solution:I'll really appreciate if anyone check the solution to tell me if it's correct or not.
DIDN'T YOU USE LATEX fonts..IF U DID THEN TRY TO USE IT PROPERLY DUDE...
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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 $
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 $
Re: Find $$(x,y)$$
hey Super Boy that's a smart way of doing it.