MHB Integer Find x,y for $y^2+2y=x^4+20x^3+104x^2+40x+2003$

[kaliprasad]

Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003$

 

[Albert1]

Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003-----(1)$

Spoiler

let $p=x^2+10x=x(x+10)-----(2)$
from (1) we have :$y(y+2)=p^2+4p+2003$
$\therefore (y+1)^2=p^2+4p+2004---(3)$
from $(2):x^2+10x-p=0---(4)$
for $x,y$ both are intgers we get :$p>0,$ and $p$ may take values from the following lists:
$11(1\times 11),24(2\times 12),39(3\times 13),----,119(7\times 17)----,n\times (n+10)$
from $(3) : p^2+4p+2004$ is a perfect square
we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
hence $(x,y)=(7,128) \, (7,-130)$
or $(x,y)=(-17,128) \, (-17,-130)$

 

[kaliprasad]

Spoiler

let $p=x^2+10x=x(x+10)-----(2)$
from (1) we have :$y(y+2)=p^2+4p+2003$
$\therefore (y+1)^2=p^2+4p+2004---(3)$
from $(2):x^2+10x-p=0---(4)$
for $x,y$ both are intgers we get :$p>0,$ and $p$ may take values from the following lists:
$11(1\times 11),24(2\times 12),39(3\times 13),----,119(7\times 17)----,n\times (n+10)$
from $(3) : p^2+4p+2004$ is a perfect square
we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
hence $(x,y)=(7,128) \, (7,-130)$
or $(x,y)=(-17,128) \, (-17,-130)$

Though the ans is right but

Spoiler

we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
is a weak assumption and some solutions could have been missing

 

[Albert1]

Though the ans is right but

Spoiler

we may assume $p^2+4p+2004=(p+10)^2\rightarrow p=119$
is a weak assumption and some solutions could have been missing

Spoiler

in fact we can set:$p^2+4p+2004=(p+k)^2$
here $k$ must be even and $2< k \leq 44$, and $p=n(n+10), n\geq 1$
and we obtain :$p=n(n+10)=\dfrac {2004-k^2}{2k-4}---(*)$
the only solution for $(*)$ is $k=10, p=119$

 

[kaliprasad]

My solution

Spoiler

Add 1 to both sides to get
$(y+1)^2 = x^4+20x^3+104x^2+40x+2004 = x^4+20x^3+ 104x^2+40x+ 4 + 2000 = (x^2+10x+2)^2 +2000$
or $(y+1)^2 - (x^2+10x+2)^2 = 2000$
for the above to have solution we need to have $(y+1)$ and $(x^2+10x+2)$ both should be even or odd and as $(x^2+10x+2) = (x+5)^2-23$ so
the $2^{nd}$ number with need to be 23 less than a perfect square.
so let us find (t,z) which are $(\pm501,\pm499),(\pm252,\pm248),(\pm129,\pm121),(\pm105,\pm95),(\pm60,\pm40),(\pm45,\pm5)$ out of which
only z = 121 which 23 less than is a perfect square
so we get
$y+1= \pm 129, (x+5) = \pm 12$ giving 4 solutions ($y=-130, x = - 17$), ($y= -130,x= 7$), ($y=128,x=7$),($y= 128, x= -17$)