An Interesting Number Theory Problem: Solve \(\displaystyle{2520}={\left({x}+{y}+{x}{y}\right)}^{{2}}+{2}{x}{y}+{2}{y}-{3}{x},{x},{y}\in{\mathbb{{Z}}}\)

Jordan Chavez

Jordan Chavez

Answered question

2022-03-17

An Interesting Number Theory Problem: Solve 2520=(x+y+xy)2+2xy+2y3x,x,yZ

Answer & Explanation

Veronica Riddle

Veronica Riddle

Beginner2022-03-18Added 9 answers

2520=x2+y2+(xy)2+2x2y+2y2x+2xy+2xy+2y3x
so, if we group in terms of y
2520=y2(x2+2x+1)+y(2x2+4x+2)+x23x
so 0=y2(x+1)2+2y(x+1)2+x23x2520
so solving the quadratic equation in the usual way:
Δ=4(x+1)44(x+1)2(x23x)=4(x+1)2(x2+2x+1x2+3x+2520)
so we have found out that
Δ=(2x+2)2(5x+2521)
but we know
y=2(x+1)2±Δ2(x+1)2=±Δ2(x+1)21=±2(x+1)5x+25212(x+1)21=±5x+2521(x+1)1
so for y to be an integer, x+1 must divide 5x+2521 and 5x+2521 must be a perfect square. Let 5x+2521=k2 so x=k225215 so
(k225215+1)2  divides  k
so (k22516)2  divides  25k
so (k22516)225|k|
which gives us k=±50,±51. If k=±50, then because x=k225215, x wouldn't be an integer so contradiction. So k=51, so x=51225215=16
Because x=16 we can now see that because
y=±5x+2521(x+1)1=±31 so y is 2 or -4.
So the solutions are (x,y)=(16,4) and (16, 2).

zakulisan337

zakulisan337

Beginner2022-03-19Added 3 answers

Denote z=x+1,w=y+1 then your result
t2+2t5x=2520(t+1)25(z1)=2519

w2z25z=2516=z(w2z5)z2516=221737
If z is even then w2z5 is odd, so
4z,z=4u,4u2w2=5u+629u1±od4, and u1737, therefore u=1,17,37,629, none of which yields a perfect square for 5u+629.
If z is odd then wz is odd, (wz)21±od4, so z1±od4,z=1,17,37,629. We have the following four cases:
z=1,w25=2516, no solution;
z=17,17w25=148w=±3x=16,y=2  or  4;
z=37,37w25=68, no solution;
z=629,629w25=4, no solution.
So the only solutions are (16, 2) and (16, -4).

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