I Love 315 to $\infty$

I have always liked to use big numbers in my Diophantine equations, specially $315$ , which is my favorite number. When you work on a equation with big numbers in it, you will end up learning a lot about the ways of bounding variables. Now we have the following equation $\color{cyan}{x^2\big(y^3+z^3\big)={\color{#D61F06}315}(xyz+7)}$

There is a constant term in the equation, which is a good point to start. Note that $x$ divides the constant term, which is $315 \times 7 = 3^2 \times 5 \times 7^2$ . If $5 | x$ then the LHS of the equation is divisible by $25$ , and it follows that $25 | 315(xyz + 7)$ or $5 | xyz + 7$ . This is false as $5 | x$ . Therefore $x$ must be in the form $3^r 7^s$ .

If $r=2$ then the LHS is divisible by $3^4$ , and, thus, $3^4 | 315(xyz + 7)$ . But $315(xyz + 7)=3^2 \times 35(3m+1)$ and the power of $3$ in $315(xyz + 7)$ will be $2$ . Therefore $r \le 1$ . Using the same reasoning, one can conclude that $s \le 1$ . Hence possible values for $x$ are $1, 3, 7, 21$ , which results in the following four equations $y^3+z^3=315(yz+7) \qquad (1) \\ y^3+z^3=35(3yz+7) \qquad (2) \\ y^3 + z^3 = 45(yz + 1) \qquad (3) \\ y^3 + z^3 = 5(3yz + 1), \qquad (4)$ respectively.

Observe that , for example, in the first equation $3|y^3+z^3$ and $5|y^3+z^3$ . In the second equation $5|y^3+z^3$ . Now we will prove for primes $p \in \left\{3, 5 \right\}$ that if $p| y^3+z^3$ then $p|y+z$ . Consider the following more general lemma:

$\color{#D61F06}\text{Lemma}$ : For every odd prime number $p$ , where $p-1 \stackrel{3}{\not\equiv} 0$ , if $p|y^3+z^3$ then $p|y+z$ .

$\color{#D61F06}\text{Proof}$ : This is clear if $p$ divides one of $y$ or $z$ , so let $p \nmid y$ and $p \nmid z$ . Then $y^{p-1} \stackrel{p}{\equiv} z^{p-1}$ by FLT . On the other hand $y^3 \stackrel{p}{\equiv} -z^3$ .

We claim that there are positive integers $a$ and $b$ such that $3a=(p-1)b+1$ . This is clear, because whatever $p$ might be (except primes that are congruent to $1$ modulo $3$ ), we can find a positive integer $b$ such that $(p-1)b+1$ is divisible by $3$ . And for an odd prime, $a$ will be an odd number.

Now using $a$ and $b$ as powers, from $y^{p-1} \stackrel{p}{\equiv} z^{p-1}$ we get $y^{(p-1)b} \stackrel{p}{\equiv} z^{(p-1)b}$ and from $y^3 \stackrel{p}{\equiv} -z^3$ we get $y^{3a} \stackrel{p}{\equiv} -z^{3a}$ . Sum two relations $y^{3a}+y^{(p-1)b} \stackrel{p}{\equiv} z^{(p-1)b}-z^{3a} \ \ \ \Rightarrow \ \ \ y^{(p-1)b}(y+1) \stackrel{p}{\equiv} z^{(p-1)b}(1-z)\stackrel{p}{\equiv} y^{(p-1)b}(1-z)$ and $y+1 \stackrel{p}{\equiv} 1-z \ \ \ \Rightarrow \ \ \ y+z \stackrel{p}{\equiv} 0$

We are going to use this lemma for decreasing coefficients and also for lowering the order of main variables in the four obtained equations.

$\boxed{\color{cyan}\text{Equation (1)}}$ Using Lemma for the equation $(1)$ , we get $y+z=15k$ , where $k$ is a positive integer and the equation becomes $15k\big(225k^2-3yz\big)=315(yz+7) \ \ \ \Rightarrow \ \ \ \frac{75k^3-49}{k+7}=yz$ Applying $(y+z)^2 \ge 4yz$ , gives $\frac{75k^3-49}{k+7}=yz \le \frac{(y+z)^2}{4}= \frac{225k^2}{4}$ Rearranging this inequality gives $75k^2(k-21) \le 196$ . It is easy to see that $k \le 21$ . Once again write $yz=\frac{75k^3-49}{k+7}=\frac{75k^3+75\times7^3-75\times7^3-49}{k+7}=75(k^2-7k+49)-\frac{2\times7^2 \times263}{k+7}$ Thus, $k+7 | 2\times7^2 \times263$ . But note that $8 \le k+7 \le 28$ and the only divisor of $2\times7^2 \times263$ that lies in this range is $14$ . Hence $k+7=14$ and $k=7$ . Therefore $(105-z)z=\frac{75\times 7^3-49}{14}=1834,$ which results in $z^2-105z+1834=0$ and no integer solutions.

$\boxed{\color{cyan}\text{Equation (2)}}$ In an exact similar manner to equation $(1)$ , one can prove that there is no integer solution for equation $(2)$ .

$\boxed{\color{cyan}\text{Equations (3) and (4)}}$ Using Lemma for the equation $(3)$ we get $y+z=15k$ and for the equation $(4)$ we get $y+z=5l$ , where $k$ and $l$ are some positive integers. Using the identity $y^3+z^3=(y+z)((y-z)^2+yz)$ equations $(3)$ and $(4)$ take the following forms: $k(y - z)^2 + (k - 3)yz = 3 \qquad (5) \\ l(y - z)^2 + (l - 3)yz = 1 \qquad (6)$ In equation $(5)$ for $y=z$ we get $(k-3)y^2=3$ which implies $y=1$ and $y+z=2 \neq 15k$ , thus $y\neq z$ and $yz \ge 2$ and $(y-z)^2 \ge 1$ . Hence if $k>3$ then LHS of $(5)$ will be greater than $3$ . It follows that $k \le 3$ . For $k=1$ , $(y-z)^2-2yz=3$ and $y+z=15$ , which fail to have integer solutions. For $k=2$ , $2(y-z)^2-yz=3$ and $y+z=30$ , which fail to have integer solutions. For $k=3$ , $(y-z)^2=1$ and $y+z=45$ , which give two solutions: $(x, y, z)=(7, 22, 23), (7, 23, 22)$ .

In equation $(6)$ , if $l>3$ then LHS of $(3)$ will be greater than $1$ . It follows that $l \le 3$ . Examining $l=1, 2, 3$ with combining equation $(6)$ and $y+z=5l$ gives two solutions again: $(x, y, z)=(21, 1, 4), (21, 4, 1)$ .

By the same arguments as Kazem, we can deduce that $x = 7,21$ .

If $x = 7$ then, putting $z = U - y$ , we obtain the equation $3(U+15)y^2- 3U(U + 15)y + (U^3 - 45) \; = \; 0$ Since this quadratic in $y$ has integer roots, its discriminant $8100 + 540 U + 2025 U^2 + 90 U^3 - 3 U^4$ must be a perfect square. Solving this equation numerically, the discriminant is only positive for integer $U$ when $-15 \le U \le 45$ , and the discriminant is a perfect square for an integer $U$ just when $U=45$ . Solving the quadratic equation for $U=45$ yields $y = 22,23$ . Thus we obtain solutions $(7,22,23)$ and $(7,23,22)$ .

Similarly, if $x=21$ then, putting $z = U -y$ , we obtain the equation $3(U+5)y^2 - 3U(U+5)y + U^3 - 5 \; = \; 0$ The discriminant $300 + 60 U + 225 U^2 + 30 U^3 - 3 U^4$ is only positive for integer $U$ when $-5 \le U \le 15$ , and the discriminant is a perfect square for integer $U$ precisely when $U=5$ . Solving the quadratic equation for $U=5$ yields $y=1,4$ . Thus we obtain solutions $(21,1,4)$ and $(21,4,1)$

Thus the solution is $7+22+23+7+23+22+21+1+4+21+4+1 = \boxed{156}$ .

It is worth noting that this method can be used to show that there are no solutions when $x=1$ or $x=3$ , so the condition that $x > 3$ could be omitted.