5. Problem: Find all pairs of integers $x, y$ such that
$$
y^{5}+2 x y=x^{2}+2 y^{4} \text {. }
$$
Solution: Rearrange and factorize to get
$$
y^{2}(y-1)\left(y^{2}-y-1\right)=(x-y)^{2} .
$$

Note that $y$ and $(y-1)$ are coprime (their greatest common divisor is 1) because they are consecutive integers. Note since $y(y-1)$ and $\left(y^{2}-y-1\right)$ are consecutive integers, we see that $\left(y^{2}-y-1\right)$ is coprime to both $y$ and $(y-1)$. Therefore the three factors
$$
y^{2},(y-1) \text { and }\left(y^{2}-y-1\right) \text { are pairwise coprime. }
$$

Since their product is a perfect square it follows that either: one of $y^{2},(y-1)$ and $\left(y^{2}-y-1\right)$ is zero, or all three of them are perfect squares. So we have four cases:
- Case A: $y=0$.
Substituting this into the original equation yields $(0)^{5}+2 x(0)=x^{2}+2(0)^{4}$. Solving this quadratic yields $x=1$ and so $(x, y)=(0,0)$ is the only solution in this case.
- Case B: $y=1$.
Substituting this into the original equation yields $(1)^{5}+2 x(1)=x^{2}+2(1)^{4}$. Solving this quadratic yields $x=1$ and so $(x, y)=(1,1)$ is the only solution in this case.
- Case C: $y^{2}-y-12$ then $(y-1)^{2}=y^{2}-2 y+1<y^{2}-y-1<y^{2}$. This would give us $(y-1)^{2}<k^{2}<y^{2}$ which is a contradiction because $(y-1)^{2}$ and $y^{2}$ are consecutive squares.
- If $y<-1$ then $(-y)^{2}<y^{2}-y-1<y^{2}-2 y+1=(-y+1)^{2}$. This would give us $y^{2}<k^{2}<(y-1)^{2}$ which is a contradiction because $(-y)^{2}$ and $(-y+1)^{2}$ are consecutive squares.
Therefore we must have $-1 \leq y \leq 2$. We have already covered $y=0$ and $y=1$ in cases A and B respectively. So it suffices now only to consider $y=-1$ and $y=2$.
- If $y=-1$ then $(-1)^{5}+2 x(-1)=x^{2}+2(-1)^{4}$. This rearranges into $(x+1)^{2}=-2$ which has no real solutions.
- If $y=2$ then $2^{5}+4 x=x^{2}+2 \times 2^{4}$. This rearranges to give $x^{2}-4 x=0$ which has solutions $x=0$ and $x=4$.
Hence in this case we have solutions $(x, y)=(0,2)$ and $(4,2)$.

In summary we have four distinct solutions for $(x, y)$ being:
$$
(x, y)=(0,0),(1,1),(0,2) \text { and }(4,2) .
$$

Alternative Solution:
In a similar manner to above get to Case D:
$$
y^{2}-y-1=k^{2} .
$$

Then rearrange it to give $(2 y-1)^{2}-4 k^{2}=5$. This then factorizes as a difference between two squares as
$$
(2 y-1+2 k)(2 y-1-2 k)=5
$$

Since 5 is prime it can only be factored in two ways: $5=5 \times 1=(-5) \times(-1)$. The sum of these two factors is $(2 y-1+2 k)+(2 y-1-2 k)=4 y-2$. Therefore:
$$
4 y-2=5+1 \quad \text { or } \quad 4 y-2=(-5)+(-1) \text {. }
$$

From $(4 y-2)=6$ we get $y=2$, and from $(4 y-2)=-6$ we get $y=-1$.
Thus we have ruled out all possibilities except for $y=-1,0,1,2$. Checking these each individually yields the four answers.
- $y=2$ yields $(2)^{5}+2 x(2)=x^{2}+2(2)^{4}$ which simplifies to become $x^{2}-4 x=0$, and has solutions $x=0$ and $x=4$.
- $y=1$ yields $(1)^{5}+2 x(1)=x^{2}+2(1)^{4}$ which simplifies to become $x^{2}-2 x+1=0$, and has solution $x=1$ only.
- $y=0$ yields $(0)^{5}+2 x(0)=x^{2}+2(0)^{4}$ which simplifies to become $x^{2}=0$, and has solution $x=0$ only.
- $y=-1$ yields $(-1)^{5}+2 x(-1)=x^{2}+2(-1)^{4}$ which simplifies to $x^{2}+2 x+3=0$, but this has no real solutions.

Thus all solutions are
$$
(x, y)=(0,2),(4,2),(1,1) \text { and }(0,0) .
$$