A student who participated in scoring(marking) the test papers has given the slip that no one had succeeded getting a single point for this problem.

Just a sketch: I would say $x = \frac{a}{b}$ and $y = \frac{c}{d}$, with $gcd(a,b) = gcd(c,d) = 1$ and then solve the diophantine equation. This gives $\frac{ac^2}{bd^2} = \frac{a^2+2ab-3b^2}{b^2}$ or $abc^2 = d^2(a+3b)(a-b)$ Now, as $gcd(a,b) = 1$, we must have $ab|d^2$ as $ab \nmid (a+3b)(a-b)$ As $gcd(c,d) = gcd(c^2,d^2)$, we even have $d^2|ab$, so therefore $d^2=ab$ and $c^2=(a+3b)(a-b)$ $c^2 + (2b)^2 = (a+b)^2$ has to be a pythagorean triplet, and $ab$ has to be a square. We can generate infintely many solutions like this if we fix $ab$ and then see if $(a+3b)(a-b)$ is indeed a square

Solution. if $y=0$, $(x, y)=(1, 0), (-3, 0)$. if $y<0$, it is equal to $y>0$. WLOG $y>0$. $xy^{2}=x^{2}+2x-3$ <=> $x^{2}+(2-y^{2})x-3=0$ This equation only has rational solutions. $D=(2-y^{2})^{2}+12=k^{2}$. ($k\in Q$) Let $y=\frac{b}{a}$. $(a, b>0, a, b\in Z, gcd(a, b)=1)$ $(2-\frac{b^{2}}{a^{2}})^{2}+12=\frac{16a^{4}-4a^{2}b^{2}+b^{4}}{a^{4}}=k^{2}$. Therefore, $16a^{4}-4a^{2}b^{2}+b^{4}=(2a)^{4}-(2a)^{2}b^{2}+b^{4}=n^{2}$. $(n\in Z)$ By Lemma., $b=2a$, $y=2$. Therefore, $(x, y)=(-1, 2), (3, 2)$. Beacause of WLOG, $(x, y)=(-1, 2), (-1, -2), (3, 2), (3, -2)$. In Conclusion, $(x, y)=(1, 0), (-3, 0), (-1, 2), (-1, -2), (3, 2), (3, -2)$.

seojun8978 wrote: A student who participated in scoring(marking) the test papers has given the slip that no one had succeeded getting a single point for this problem. I'm a Korean and I took this test. It's right. I just knew the solution when the test finished...

Mathlion1212 wrote: Lemma. $x^{4}-x^{2}y^{2}+y^{4}=z^{2}$'s solution is $(x, y, z)=(n, n, n^{2})$. Can you give a proof for this?

faraday_vij wrote: Mathlion1212 wrote: Lemma. $x^{4}-x^{2}y^{2}+y^{4}=z^{2}$'s solution is $(x, y, z)=(n, n, n^{2})$. Can you give a proof for this? I know the solution but it is not that trivial I am on the phone now so I cannot post it I will post it by tuesday

Lemma> $x^4 - x^2y^2 + y^4 = z^2$ satisfies if and only if $(x,y,z) = (k,k,k^2)$ or $(0,k,k^2)$ or $(k,0,k^2)$ Proof> if $xy = 0, (x,y,z) = (k,0,k^2)$ or $(0,k,k^2)$ and if $x=y$, $(x,y,z) = (k,k,k^2)$ Assuming that $xy >0$ and $x \neq y$ Let $(x,y,z)$is the solution such that $z$ is the minimum of all the integer solutions if a prime $p$ such that $p|x, p|y$ exists, $\left(\frac{x}{p},\frac{y}{p},\frac{z}{p^2}\right)$is also a solution which is contradictory to the minimality of $z$ Therefore, $gcd(x,y)=1$ $x^4 -x^2y^2 + y^4 = z^2 \Leftrightarrow (x^2 - y^2)^2 + (xy)^2 = z^2$ $gcd(x^2 - y^2,xy) = 1$ ⅰ)$x,y$ are all odd numbers By Pythagorean Triplet, $a,b \in \mathbb{N}$ such that $x^2 - y^2 = 2ab, xy = a^2 - b^2,\,\, gcd(a,b) = 1$ exists $a^4 - a^2b^2 + b^4 = \left(\frac{x^2+y^2}{2}\right)^2$ Therefore, $\left(a,b,\frac{x^2+y^2}{2}\right)$ also is a solution $z = (a^2+b^2)^2 > a^4 - a^2b^2 + b^4 = \left(\frac{x^2+y^2}{2}\right)^2$ contradiction ⅱ)There is an even number in ${x,y}$ $WLOG$ $2|x$ Let $x = 2x_1$ By Pythagorean Triplet, $a,b \in \mathbb{N}$ such that $4x_1^2 - y^2 = a^2 - b^2, 2x_1y = 2ab, \,\, gcd(a,b) = 1$ exists From $x_1y = ab$, positive integers $m,n,p,q$ such that $x_1 = mn, y = pq, a = mp, b = nq$ exists Then $4m^2n^2 - p^2q^2 = m^2p^2 - n^2q^2 \Leftrightarrow n^2(4m^2+q^2) = p^2(m^2 + q^2)$ Since $gcd(x_1,y) = 1, gcd(m,q) = gcd(n,p) = 1$ Since $m^2 + q^2 \not\equiv 0 (mod 3), gcd(4m^2+q^2,m^2+q^2) = 1$ $\therefore m^2 +q^2 = n^2, 4m^2 + q^2 = p^2$ if $2|q$ then $2|p$, from$m^2 + q^2 = n^2,$ By Pythagorean Triplet, positive integers $u,v$ such that $m=u^2 - v^2, q = 2uv$ and satisfies $u^4 - u^2v^2+v^4 = \left(\frac{p}{2}\right)^2$ $\therefore \left(u,v,\frac{p}{2}\right)$ also is a solution $z = a^2 + b^2 > a = mp \ge p > \frac{p}{2}$ contradiction if $q$ is an odd number By Pythagorean Triplet from $(2m)^2 + q^2 = p^2$, positive integers $u,v$ such that $2m = 2uv, q = u^2 - v^2$ exists $u,v$ satisfies $u^4 - u^2v^2 + v^4 = n^2$ $\therefore (u,v,n)$ also is a solution $z = a^2 + b^2 > b = nq \ge n$ contradiction Hence, there are no more solutions $\therefore$$x^4 - x^2y^2 + y^4 = z^2$ satisfies if and only if $(x,y,z) = (k,k,k^2)$ or $(0,k,k^2)$ or $(k,0,k^2)$