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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Does this correct? Please advice. Thanks in advanced.
Best Answer
The second attempt works when you render the modulo $7$ results correctly:
$x^3\in\{0,1,6\}$
$2^y\in\{1,2,4\}\implies2^y+15\in\{2,3,5\}$.
$\rightarrow\leftarrow$