(xy-7)^2=(x^2+y^2)
indian math olympiad
hint:factorization you all need to hang about
diophantine equation
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determine all non negative integers (x,y) for which the relation hold
(xy-7)^2=(x^2+y^2)
indian math olympiad
hint:factorization you all need to hang about
(xy-7)^2=(x^2+y^2)
indian math olympiad
hint:factorization you all need to hang about
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: diophantine equation
$\begin{align*}
& (xy-7)^2 =(x^2+y^2) \\
\iff & x^2y^2-14xy+49 =x^2+y^2\\
\iff & x^2y^2-12xy+36+13 =x^2+2xy+y^2\\
\iff & (xy-6)^2+13 =(x+y)^2 \\
\iff & (x+y-xy+6)(x+y+xy-6) =13
\end{align*}$
Now using another factoring trick $(ab+a+b+1)=(a+1)(b+1)$, we shall get a few cases (as $13$ is a prime). Checking the cases we can find the answers.
& (xy-7)^2 =(x^2+y^2) \\
\iff & x^2y^2-14xy+49 =x^2+y^2\\
\iff & x^2y^2-12xy+36+13 =x^2+2xy+y^2\\
\iff & (xy-6)^2+13 =(x+y)^2 \\
\iff & (x+y-xy+6)(x+y+xy-6) =13
\end{align*}$
Now using another factoring trick $(ab+a+b+1)=(a+1)(b+1)$, we shall get a few cases (as $13$ is a prime). Checking the cases we can find the answers.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
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Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Re: diophantine equation
it can also be done in the following way.
since,$(xy-7)^2=x^2+y^2$
It can be the sides of a right-angled triangle.Now the formula for determining the sides of a triangle which are whole numbers is $(1+2n)^2+(2n(n+1))^2=(1+2n(n+1))^2$.
Now, by solving $(xy-7)^2=(1+2n(n+1))^2$,
$x^2=(1+2n)^2$ and
$y^2=(2n(n+1))^2$.
We get the values of x and y.
$(x,y)=(3,4)$
therefore,$(x,y)=(4,3)$ is another solution.
Again,by solving $(xy-7)^2=k^2(1+2n(n+1))^2$,
$x^2=k^2(1+2n)^2$ and
$y^2=k^2(2n(n+1))^2$.
We get the values of x and y.
But x and y there has no solution.
Therefore,the two solutions are $(x,y)=(3,4)$ and $(x,y)=(4,3)$.
since,$(xy-7)^2=x^2+y^2$
It can be the sides of a right-angled triangle.Now the formula for determining the sides of a triangle which are whole numbers is $(1+2n)^2+(2n(n+1))^2=(1+2n(n+1))^2$.
Now, by solving $(xy-7)^2=(1+2n(n+1))^2$,
$x^2=(1+2n)^2$ and
$y^2=(2n(n+1))^2$.
We get the values of x and y.
$(x,y)=(3,4)$
therefore,$(x,y)=(4,3)$ is another solution.
Again,by solving $(xy-7)^2=k^2(1+2n(n+1))^2$,
$x^2=k^2(1+2n)^2$ and
$y^2=k^2(2n(n+1))^2$.
We get the values of x and y.
But x and y there has no solution.
Therefore,the two solutions are $(x,y)=(3,4)$ and $(x,y)=(4,3)$.
Re: diophantine equation
missing.........non-negetiv integr......(0,7);(7,0)