# diophantine equation $x^3+x^2-16=2^y$

Solve in integers: $x^3+x^2-16=2^y$.

my attempt:

of course $y\ge 0$, then $2^y\ge 1$, so $x\ge 1$.

for $y=0,1,2,3$ there is no good $x$.

so $y\ge 4$ and we have equation $x^2(x+1)=16(2^z+1)$, where $z=y-4\ge 0$.

what now?

• A quick search shows $x,y = 4,6$ is the only solution with $0<y<100$. Apr 5, 2015 at 0:53
• can somebody solve this for odd values of x?
Apr 17, 2015 at 15:02
• using parity doesn't work, i think
Apr 17, 2015 at 15:11
• @tong_nor What about this? $$x^3 +x^2 = 2^y+2^4 \implies x^2(x+1) = 2^4(2^{y-4} +1) \implies x^2(x+1) = 4^2(4^{0.5y-2} +1)$$ I am not sure but can we conclude due to some reason which I may have overlooked that $x$ can't take more than one values here so we can find $y$ and find the solution. I feel that x can't take more than one values. The structure of LHS and RHS is very alike. Mar 4, 2016 at 5:07

$x^3+x^2=2^y+16$. The RHS is positive, so $x^2(x+1)>0\iff x\ge 1$. Since $2^y$ is an integer, we have $y$ is a positive integer too ($y=0$ won't give a solution).

$x,y$ are positive integers.

$x^3+x^2-16=2^y$. You see a cubic polynomial on the LHS that could easily be strictly bounded between two consecutive cubes (namely $x^3$ and $(x+1)^3$) for most values of $x$, making it impossible for it to be a cube itself. So if $2^y$ is a cube, i.e. $3\mid y$, we're done. And indeed it is a cube.

$2^y\equiv 1,2,4\pmod{7}$ for $y\equiv0,1,2\pmod 3$, respectively.

$x^3+x^2-16\equiv 5,0,3,6,1,1,5\pmod{7}$ for $x\equiv 0,1,2,3,4,5,6\pmod{7}$, respectively.

The only common residue is $1$, so $y\equiv 0\pmod{3}$. This implies $x^3+x^2-16$ is a cube.

But $x^3<x^3+x^2-16<(x+1)^3,\forall x\ge 5$, so $1\le x\le 4$, which only give $(x,y)=(4,6)$.

• How did you know to do it mod 7?
Apr 18, 2015 at 17:18
• i don't understand how you got this line: 2^y≡1,2,4(mod7) for y≡0,1,2(mod3)
Apr 18, 2015 at 17:25
• @user3491648 You work logically in the route of proving that $3\mid y$. You want $2^y\bmod n$ to have a period 3. $2^y=1,2,4,8,\ldots$, so $8\equiv 1\pmod {n}\Leftrightarrow n=7$. Or you let $y=3k+r$ ($0\le r\le 2$) and want to prove that $r=0$. $2^{3k+r}=8^k2^r$ and $8^k\equiv 1\pmod {n}\Leftrightarrow n=7$ (we want it to be independent of $k$). Apr 18, 2015 at 17:32
• what $y$ is mod 6 tells you what $2^y$ is mod 7. but we only know what $y$ is mod 3.
Apr 18, 2015 at 17:32
• @user3491648 As for your 2nd comment, if we let $y=3k+r$ ($0\le r\le 2$), then $2^y\equiv 8^k2^r\equiv 2^r\pmod{7}$, which is independent of $k$ and $r=0,1,2$ give $2^r\equiv 1,2,4\pmod{7}$, respectively. Or you can simply find the pattern in $2^0\equiv 1, 2^1\equiv 2,\ 2^2\equiv 4, 2^3\equiv 1, 2^4\equiv 2, 2^5\equiv 4 \pmod 7$. If you don't find this rigorous enough, you can use induction -- $2^k\equiv 1, 2^{k+1}\equiv 2, 2^{k+2}\equiv 4\Rightarrow 2^{k+3}\equiv 8\cdot 2^k\equiv 1, 2^{k+4}\equiv 8\cdot 2^{k+1}\equiv 2, 2^{k+5}\equiv 8\cdot 2^{k+2}\equiv 4\pmod {7}$. Apr 18, 2015 at 17:36

This is only a solution for even values of $x$:

For $x$ even, $x^2$ is even and $x+1$ is odd. So $\gcd(16, x+1) = 1$. So $16 \mid x^2$. So $4 \mid x$. So $x = 4k$ for some $k$.

$16k^2(4k+1)=16(2^z+1)$

$k^2(4k+1)=2^z+1$

If $k$ is even then the RHS is even and we have a contradiction. So $k$ must be odd. So $k = 2t+1$ for some $t$.

$(2t+1)^2 (8t+5) = 2^z + 1$

$(4t^2 + 4t + 1)(8t + 5) = 2^z + 1$

$32t^3 + 52t^2 + 28t + 4 = 2^z$

We see $z \geq 2$. Let $w = z - 2$.

$8t^3 + 13t^2 + 7t + 1 = 2^w$

For $w \geq 1$, $t^2 + t + 1 \equiv 0 \pmod{2}$ which has no solution in $t$.

So $w = 0$.

$8t^3 + 13t^2 + 7t = 0$

$t(8t^2 + 13t + 7) = 0$

Take the discriminant of the quadratic, so we get $169 - 224 < 0$.

So $t = 0$ and $w = 0$.

So for $x$ even, $x = 4k = 4(2t + 1) = 4$ and $y = z + 4 = w + 6 = 6$.

You can also bound (bind?) $x$, giving it in terms of $y$. Go back to $x^3+x^2−16=2^y$. Consider $f(x,y) = x^3 + x^2 - 16 - 2^y$. Evaluate $f(2^{y/3},y) = 2^y + 2^{2y/3} - 16 - 2^y = 2^{2y/3} - 16$ which equals $0$ when $y = 6$ and is bigger than $0$ when $y > 6$. Evaluate $f(2^{y/3} - 1, y)$ which you can check here is less than $0$ for all $y$. This shows that if $x$ is an integer and $f(x,y) = 0$ then $x = \left \lfloor 2^{y/3} \right \rfloor$.

As far as working with $x = \left \lfloor 2^{y/3} \right \rfloor$, you can write it as $x = \left \lfloor 2^q2^{r/3} \right \rfloor$ where $0 \leq r < 3$ which can be expressed as $\left \lfloor 10_2^q 10_2^{r/3} \right \rfloor$ in binary. As $q$ increases (which increases $y$ by 3), the value of $x$ either gets doubled, or is double plus one. I'm not sure if that's useful.

How to finish this?

• What about this? $$x^3 +x^2 = 2^y+2^4 \implies x^2(x+1) = 2^4(2^{y-4} +1) \implies x^2(x+1) = 4^2(4^{0.5y-2} +1)$$ I am not sure but can we conclude due to some reason which I may have overlooked that $x$ can't take more than one values here so we can find $y$ and find the solution. I feel that x can't take more than one values. The structure of LHS and RHS is very alike. Mar 4, 2016 at 5:10

$x^2(x+1) = 2^y + 16$

Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.

$$\begin{array}{n|c|c|} \hline n & n^2(n+1) & 2^n + 16 \\ \hline 0 & 0 & 17 \\ 1 & 2 & 18 \\ 2 & 12 & 20 \\ 3 & 36 & 24 \\ 4 & 80 & 32 \\ 5 & 150 & 48 \\ 6 & 252 & 80\\ \hline \end{array}$$

Note that the table indicates that $(x,y) = (4,6)$ is a solution.

So any solution involving $y \ge 7$ will require $x \ge 5$.

We will show that there is no solution for $y \ge 7$.
So we can assume now that $x \ge 5$ and $y \ge 7$.

\begin{align} x^2(x+1) &= 2^y + 16\\ x^2(x+1) &= 16(2^{y-4} + 1)\\ \end{align}

Note that $2^{y-4}+1$ must be an odd integer.

So if $x$ is an odd integer, $\gcd(x+1,2^{y-1}+1) = 1$.
$\quad$ Hence $x+1 | 16$
$\quad$ Remembering that $x \ge 5$, we must have $x = 7$ or $x = 15$.
$\quad$Case $1: x = 7$
\begin{align} 16(2^{y-4}+1) &= 392\\ 2(2^{y-4}+1) &= 49 & \text{Has no solution.}\\ \end{align}
$\quad$Case $2: x = 15$
\begin{align} 16(2^{y-4}+1) &= 3600\\ 2^{y-4}+1 &= 225 & \\ 2^{y-4} &= 224 \\ 2^{y-4} &= 32 \times 7 & \text{Has no solution.}\\ \end{align}

So if $x$ is an even integer, $\gcd(x^2,2^{y-1}+1) = 1$.
$\quad$ So $x^2 | 16$. This can't happen since $x \ge 5$.