4
$\begingroup$

Show that there are no two positive integers $x$ and $y$ such that $x^3=2^y+15$.

Attempt

For the sake of contradiction, suppose that there are two positive integers $x$ and $y$ satisfying $x^3 = 2^y + 15$.

Consider the equation modulo $4$. The cubes of integers modulo $4$ can only yield remainders of $0$, $1$, or $3$. This can be verified by calculating the cubes of the numbers $0$, $1$, $2$, and $3$ modulo $4$.

Now, let's analyze the possible remainders of powers of $2$ modulo $4$. We have $2^0 \equiv 1 \pmod{4}$, $2^1 \equiv 2 \pmod{4}$, $2^2 \equiv 0 \pmod{4}$, and $2^3 \equiv 0 \pmod{4}$. As we can see, for $y \geq 2$, $2^y$ is divisible by $4$.

Using this information, let's consider the equation $x^3 = 2^y + 15$ modulo $4$. We have two cases:

Case 1: $y = 1$ If $y = 1$, then $2^y = 2$, and the equation becomes $x^3 = 2 + 15$. This simplifies to $x^3 = 17$. However, no positive integer cubed equals $17$. This contradicts the equation, so this case is not possible.

Case 2: $y \geq 2$ If $y \geq 2$, then $2^y$ is divisible by $4$. Adding $15$ to $2^y$ yields a number that leaves a remainder of $3$ when divided by $4$. However, as mentioned earlier, the cubes of integers modulo $4$ can only yield remainders of $0$, $1$, or $3$. Therefore, there are no positive integers $x$ and $y$ that satisfy the equation $x^3 = 2^y + 15$ in this case.

Since we have considered all possible cases and found contradictions in each case, we can conclude that there are no positive integers $x$ and $y$ that satisfy the equation $x^3 = 2^y + 15$. Thus, the assumption that such integers exist must be false.

Hence, we have proven by contradiction that there are no two positive integers $x$ and $y$ satisfying the equation $x^3 = 2^y + 15$. Q.E.D.

${}$

${}$

Second Attempt:

Suppose, for the sake of contradiction, that there are two positive integers $x$ and $y$ such that $x^3=2^y+15$.

Notice that for any positive integer $x$, we have $x^3 \equiv 0,1,6 \pmod 7$.

On the other hand, for any positive integer $y$, we have $$2^y \equiv 1,2,4 \pmod 7, \tag{1}$$ i.e.,$2^y+15 \equiv 2,3,5 \pmod 7$.

Comparing the congruence relations, we see that there is no overlap between the possible values of the left side ($0,1,6$) and the possible values of the right side ($2,3,5$) modulo $7$.

Hence, our assumption is false, which means that there are no two positive integers $x$ and $y$ such that $x^3=2^y+15$. Q.E.D.

${}$

Does this correct? Please advice. Thanks in advanced.

$\endgroup$
9
  • 1
    $\begingroup$ Where do have doubts? You already had this comment in your last question: For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. $\endgroup$ May 23, 2023 at 8:49
  • 2
    $\begingroup$ In case 2) you say that $15+2^y$ has remainder $3$, then say that $3$ is also a possible remainder of the cube. $\endgroup$ May 23, 2023 at 8:50
  • 1
    $\begingroup$ Shouldn't $2^y+15\equiv2,3,5\mod7$ instead of $0,2,5$? But that arrives at the conclusion even faster (since $x^3\equiv0,1,6\mod7$ doesn't match at all), good job 🙂 $\endgroup$ May 23, 2023 at 9:49
  • 1
    $\begingroup$ @TheMather-orratherAMather Ah, my bad. Thanks for pointing out my mistake. So, the proof was correct? $\endgroup$
    – math404
    May 23, 2023 at 9:58
  • 1
    $\begingroup$ @math404 yes, it's correct👌 $\endgroup$ May 23, 2023 at 10:04

2 Answers 2

3
$\begingroup$

The second attempt works when you render the modulo $7$ results correctly:

$x^3\in\{0,1,6\}$

$2^y\in\{1,2,4\}\implies2^y+15\in\{2,3,5\}$.

$\rightarrow\leftarrow$

$\endgroup$
1
$\begingroup$

Case 2 is handled incorrectly. You basically say:

  1. If $y\geq 2$, then $2^y+15$ has remainder $3$ when divided by $4$.
  2. Any cube must have remainder $0,1$ or $3$ when divided by $4$.
  3. Therefore, $2^y+15$ is not a cube.

But the mistake there is, quite simply, that the conclusion (3) does not follow from the premises (1 and 2).

$\endgroup$
2
  • $\begingroup$ So, how to handle it? $\endgroup$
    – math404
    May 23, 2023 at 8:57
  • 1
    $\begingroup$ what about my second attempt? $\endgroup$
    – math404
    May 23, 2023 at 9:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .