A Diophantine Equation | Integer Solutions

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PLAYLISTS 🎵 :

[dwm]

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.

[fqgugmz]

（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)

[nasrullahhusnan]

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)}

[mcwulf]

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

[neuralwarp]

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

[davidtaran]

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. x<0 y>0
k=-x>0
y=(17+k)/(1-2k)<0
3.x<0 y<0
k=-x>0
y=(17+k)/(1-2k)<0

3.1.
k=1, x=-1, y=-18
x=-18, y=-1
3.2.
k=3, x=-3, y=-4
k=3, x=-4, y=-3
3.3.
k=4, x=-4, y=-3 -> rpt
3.4.
k=18, x=-18, y=-1 ->rpt
1. x>0, y<0
y=(17-x)/(1+2x)
y=0.5*(33/(1+2x)-1)<0 ->haven't negative int roots

[branialtocci]

I'm your video's 5³ th liker XD