A Diophantine Equation | Integer Solutions

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[@dwm1943]

I got the 4 (or 8) solutions quickly enough, by an odd and even argument, plus a bit of trial and error. I was happy that these were the only solutions, but your elegant method 1 proved it rigorously. Lovely little factorising trick. Made me smile.

[@miguelaphan58]

..fascinating

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@ViaanHossain]

Very nice

[@SyberMath]

Thank you!

[@赖皮球]

（2x+1)(2y+1)=17*2+1=35 (2x+1,2y+1) (-1,-35 ; 1,35 ; -5,-7 ; -7,-5 ; -35,-1 ; 35 ,1 ; 7,5 ;5,7)

[@nasrullahhusnan2289]

Note the equation is cyclical w.o.l.g. assume that x>y Trivial solution if y=0 --> x=17 x(1+2y)+½(2y+1)=17+½ (2y+1)(2x+1)=35 2x+1=7 --> x=3 2y+1=5 --> y=2 Thus (x,y)={(17,0), (0,17), (3,2), (2,3)}

[@Generalist18]

Man I missed trivial solution

[@giuseppemalaguti435]

x(1+2y)=17-y...x=(17-y)/(1+2y)..y=2(x=3)..y=-3(x=-4)

[@mcwulf25]

I started with #1 but then doubled both sides rather than work with the 1/2.

[@SyberMath]

Good thinking

[@Ahwke]

y=(17-x)/(2x+1)😊

[@davidtaran952]

before watching the video: x+y+2xy=17 symmetrical equation: if {x=t1,y=t2} is a root, then {x=t2,y=t1} is a root too. x+y(1+2x)=17 y=(17-x)/(1+2x) 1. x>=0, y>=0 1.1. x=0 y=17 x=17 y=0 1.2. x=2 y=3 y=2 x=3 2. x0 k=-x>0 y=(17+k)/(1-2k)0, y

[@branialtocci]

I'm your video's 5³ th liker XD

[@SyberMath]

Nice! Thanks a 10^6 😃

[@roberttelarket4934]

@@SyberMath: Ha ha 10^6 very clever!

[@neuralwarp]

By the vey, the letter W is the liquid form of U and never denotes a V sound.

[@SyberMath]

😀

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Welcome to my skibidi toilet

[@Rosiesmathchannel]

Wut