No integer whose digits add up to ## 15 ## can be a square or a cube

[Math100]

Homework Statement

Prove that no integer whose digits add up to $15$ can be a square or a cube.
[Hint: For any $a$, $a^{3}\equiv 0, 1,$ or $8\pmod {9}$.]

Relevant Equations

None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7$, or $8\pmod {9}$.
This means $a^{2}\equiv 0, 1, 4, 9, 7, 7, 0, 4$, or $1\pmod {9}$ and $a^{3}\equiv 0, 1, 8, 0, 1, 8, 0, 1$, or $8\pmod {9}$.
Thus $a^{2}\equiv 0, 1, 4$, or $7\pmod {9}$ and $a^{3}\equiv 0, 1$, or $8\pmod {9}$.
There exists no integer $a$ such that $a^{2}\equiv 6\pmod {9}$ or $a^{3}\equiv 6\pmod {9}$.
Therefore, no integer whose digits add up to $15$ can be a square or a cube.

 

[fresh_42]

Homework Statement:: Prove that no integer whose digits add up to $15$ can be a square or a cube.
[Hint: For any $a$, $a^{3}\equiv 0, 1,$ or $8\pmod {9}$.]
Relevant Equations:: None.

Proof:

Let $a$ be any integer.
Then $a\equiv 0, 1, 2, 3, 4, 5, 6, 7$, or $8\pmod {9}$.
This means $a^{2}\equiv 0, 1, 4, 9, 7, 7, 0, 4$, or $1\pmod {9}$ and $a^{3}\equiv 0, 1, 8, 0, 1, 8, 0, 1$, or $8\pmod {9}$.
Thus $a^{2}\equiv 0, 1, 4$, or $7\pmod {9}$ and $a^{3}\equiv 0, 1$, or $8\pmod {9}$.
There exists no integer $a$ such that $a^{2}\equiv 6\pmod {9}$ or $a^{3}\equiv 6\pmod {9}$.
Therefore, no integer whose digits add up to $15$ can be a square or a cube.

Let's see:
\begin{align*}
N&=a_k\cdot 10^k+\ldots +a_1\cdot 10 +a_0 = a^m \text{ with } m\in \{2,3\} \text{ and } \sum a_i =15\\
&\Longrightarrow \\
N&\equiv a_k \cdot 1^k +\ldots +a_1\cdot 1 +a_0 \equiv 15 \equiv 6 \pmod{9}
\end{align*}

Your solution is correct, I just needed $10^j\equiv 1\pmod 9$ to close the gap between $\sum a_i =15$ and $a^m \not\equiv 6 \pmod 9.$

 

[pbuk]

Your solution is correct, I just needed $10^j\equiv 1\pmod 9$ to close the gap between $\sum a_i =15$ and $a^m \not\equiv 6 \pmod 9.$

But that is the only non-trivial step! Without it the 'proof' is worthless.

 

[Delta2]

But that is the only non-trivial step! Without it the 'proof' is worthless.

The OP has done many problems here where he states that a number modulo 9 is equal to the sum of its digits (in base 10) modulo 9 so in his mind he has this step self implied I think. But of course the objective reader can't be inside his mind to know what he is thinking.

 

[pbuk]

The OP has done many problems here where he states that a number modulo 9 is equal to the sum of its digits (in base 10) modulo 9 so in his mind he has this step self implied I think. But of course the objective reader can't be inside his mind to know what he is thinking.

Indeed. And even if the OP decides to assume this without proof, they still need to state that they are assuming this without proof. The same goes for the assumption that $a^2 \pmod 9 \equiv (a \pmod 9)^2 \pmod 9$ at the beginning.

Why the OP doesn't think it is important to include these assumptions but thinks that it is important to state that $a \in \mathbb N \Rightarrow a\equiv 0, 1, 2, 3, 4, 5, 6, 7 \text{ or } 8 \pmod 9$ is unfathomable to me.