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

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.

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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.

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• 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. May 23, 2023 at 8:49
• In case 2) you say that $15+2^y$ has remainder $3$, then say that $3$ is also a possible remainder of the cube. May 23, 2023 at 8:50
• 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 ðŸ™‚ May 23, 2023 at 9:49
• @TheMather-orratherAMather Ah, my bad. Thanks for pointing out my mistake. So, the proof was correct? May 23, 2023 at 9:58
• @math404 yes, it's correctðŸ‘Œ May 23, 2023 at 10:04

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

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

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

$$\rightarrow\leftarrow$$

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).

• So, how to handle it? May 23, 2023 at 8:57
• what about my second attempt? May 23, 2023 at 9:45