Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]
dduclam wrote: Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]
martosss wrote:dduclam wrote: Solve for integers (x,y) : [tex]4x^2+y^2+5xy+1=3(3x+y)[/tex]
Let us examine the equation towards y:
[tex]y^2+(5x-3)y+4x^2-9x+1=0[/tex]
[tex]D=25x^2-30x+9-16x^2+36x-4=9x^2+6x+5=(3x+1)^2+4[/tex]
Now [tex]D[/tex] should be an integer, because [tex]y[/tex] is an integer, so [tex](3x+1)^2+4=k^2,\: x,k\in Z[/tex] and I think this has no integer solutions, because the distance between the integer exact squares grows constantly with 2
[tex]0^2=0,[/tex]
[tex]1^2=1,[/tex]
[tex]1-0=1, 1-0=1[/tex]
[tex]2^2=4, 4-1=3 = 1+2[/tex]
[tex]3^2=9, 9-4=5 = 3+2 e.t.c.[/tex],
so the only solution is when [tex](3x+1)^2=0,\: that\: is\: x=-\frac{1}{3}\notin Z[/tex]
That shows that there are no such x and y
Guest wrote:hi ,
how we will use latex in this forum ................Please post some solutions .
[tex][/tex]
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