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As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.

 

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

 

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$

As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$

As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$

As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$

As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to this trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution goes as follows:

We consider the more general eq.

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$

As I was browsing through the problems & solutions department of the American Mathematical Monthly the other day, I found out that, in his solution to problem E-2332 [1972, 87], which was published on page 77 of the first issue of vol. 80 of the AMM, Ernst Trost resorted to the trick and, what is more, added a reference to the aforementioned paper of his for further applications of it.

The problem was posed by R. S. Luthar and asked to find all solutions in positive integers of the equation $y^{3}+4y= z^{2}$. Trost's solution went as follows:

We consider the more general eq.

$$ay^{3} +4a^{3}b^{4}y= z^{2}, \quad y, z >0$$

where $a, b \in \mathbb{N}$. If $(y,z)$ is a solution, then the quadratic polynomial $$P(t) =ay^{3}t^{2} -z^{2}t+4a^{3}b^{4}y$$ has the rational zero $t=1$. Therefore, $z^{4}-(2aby)^{4}$ must be the square of an integer. Taking into account that $y>0$ and applying a result of Fermat (see, for instance: K. Conrad, "Proofs by descent", p. 7), we infer that $z=2aby$; it follows that the unique solution of the equation under consideration is $$(y,z)= (2ab^{2},4a^{2}b^{3}).$$