If $x,\,y$ be integers and suppose that $x^{2}- y^{3}= 17$, prove each following proposition true.

diophantine equationsinequalitynumber theoryprime numbersproof-verification

Problem. If $x,\,y$ be integers and suppose that $x^{2}- y^{3}= 17$, prove each following proposition true :

  • $y^{2}+ 2\,x+ 2$ is a prime number if and only if $\{\!x= 4,\,y= -\,1\!\}$
  • $x^{2}+ y^{2}+ x+ y$ is a composite number
  • $2\,x^{2}+ y^{2}$ is a composite number
  • $y^{2}+ 2\,x$ is not a prime number

Remark. In 1930, Trygve Nagel proved that $x^{2}- y^{3}= 17$ only has the following integer solutions
$$y= -\,2,\,-\,1,\,2,\,4,\,8,\,43,\,52,\,5234$$

  • Remark: For a., we have from the hypothesis: $(x- 4)(x+ 4)= (y+ 1)(y^{2}- y+ 1)$, we let :
    $$p_{1}(x- 4)= (p_{2}- p_{1})(y+ 1)$$
    $$(p_{2}- p_{1})(x+ 4)= p_{1}(t- y+ 1)$$
    $$t+ 2\,x+ 2= p_{3}$$

for $p_{1},\,p_{2}- p_{1},\,p_{3}\in \mathbb{Z}$ and $p_{2}\neq p_{1}$ and $t= y^{2}$. Of course, we'll solve the system of above equations
$$x+ y= \frac{-\,5\,p_{2}+ p_{1}p_{3}}{p_{2}}$$
$$y= \frac{-\,8\,p_{1}p_{2}+ p_{1}^{2}p_{3}- p_{2}^{2}}{p_{2}^{2}}$$
Therefore, $p_{2}\mid p_{1}p_{3}$ and $p_{2}^{2}\mid -\,8\,p_{1}p_{2}+ p_{1}^{2}p_{3}$. Again, we have $p_{2}= -\,p_{1}$ or $p_{2}= p_{3}$ or $p_{2}= -\,p_{3}$ .

  • For $p_{2}= -\,p_{1}$, and $t= y^{2}$, we also have that $p_{3}^{2}+ 9\,p_{3}+ 27= 0$ clearly has no integer solution.

  • For $p_{2}= p_{3}$, and $t= y^{2}$, the Diophantine equation
    $$32\,p_{1}p_{3}- 4\,p_{1}^{2}p_{3}- 5\,p_{3}^{2}+ 2\,p_{3}^{2}p_{1}- p_{3}^{3}+ 64\,p_{1}^{2}- 16\,p_{1}^{3}+ p_{1}^{4}= 0$$
    only has 2 solutions $p_{1}= 2$ and $p_{1}= -\,2$. After that, we'll doublecheck it, find $\{\!x= 4,\,y= -\,1\!\}$

  • For $p_{2}= p_{3}$, and $t= y^{2}$, the Diophantine equation
    $$-\,32\,p_{1}p_{3}- 4\,p_{1}^{2}p_{3}- 5\,p_{3}^{2}- 2\,p_{3}^{2}p_{1}- p_{3}^{3}+ 64\,p_{1}^{2}+ 16\,p_{1}^{3}+ p_{1}^{4}= 0$$
    only has 2 solutions $p_{1}= 2$ and $p_{1}= -\,2$, that's as same as For. $p_{2}= p_{3}$, find $\{\!x= 4,\,y= -\,1\!\}$

  • For b., $x\geqq 0$ or $x\leqq -\,1$, $y\geqq 0$ or $y\leqq -\,1$ then $x^{2}+ x+ y^{2}+ y= x(x+ 1)+ y(y+ 1)\geqq 0$

  • So $x^{2}+ y^{2}+ x+ y$ is a composite number because $2\mid x^{2}+ y^{2}+ x+ y$. We have the $\lceil$ q.e.d ! $\rfloor$

  • For c., my idea is to use $\mod 2$ and $\mod 3$ here. I will post a solution for this in a soon time .

  • For d., I think that d. is as same as a. partly, but I still have no idea for this. I need to the helps !

Two of all the aboves given a solution, and I'm looking forward to seeing a nicer one(s), thank you !

Best Answer

For c. (as promised): There are 2 cases: For $3\nmid x\,\therefore\,3\nmid y\,\therefore\,3\mid 2\,x^{2}+ y^{2}$ and for $3\mid x$ (2 cases in):

  1. For $2\nmid x\,\therefore\,2\mid 2\,x^{2}+ y^{2}$

  2. For $2\mid x\,\therefore\,6\mid x\,\therefore\,y^{3}\equiv -\,17(\!\mod 36\!)\,\therefore\,y^{3}\equiv 7(\!\mod 12\!)\,\therefore\,y= 12\,b+ 7\,(\!b\in \mathbb{Z}\!)$. So :

    a. For $x= 12\,a+ 6\,(\!a\in \mathbb{Z}\!)\,\therefore\,(12\,b+ 7)^{3}+ 17= (12\,a+ 6)^{2}\,\therefore\,84\,b^{2}+ 7\,b+ 10\equiv 1(\!\mod 8\!)$. So $$\therefore\,b\equiv 5(\!\mod 8\!)\,\therefore\,y= 96\,w+ 67\,(\!w\in \mathbb{Z}\!)$$ $$\therefore\,\left.\begin{matrix} y^{3}+ 17\equiv 3^{3}+ 17\equiv 44\equiv 12(\!\mod 32\!)\\ x^{2}\equiv 0,\,4,\,16(\!\mod 32\!) \end{matrix}\right\}\,\therefore\,(\!impossible\,!\!)$$ b. For $x= 12\,a\,\therefore\,(12\,b+ 7)^{3}+ 17= (12\,a)^{2}$ $$\therefore\,a^{2}\equiv 6(\!\mod 7\!)\,\therefore\,(\!impossible !\!)\,\because\,4\,a^{2}\equiv 0,\,1,\,2,\,4(\!\mod 7\!)$$ q.e.d

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