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Verify that if an integer is simultaneously a square and a cube, then it must be either of the form ${7k}$ or ${7k +1}$.

I have no idea on how to proceed.

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4 Answers 4

up vote 14 down vote accepted

If $n$ is simultaneously a cube,

$$n\equiv a^3\equiv -1,0,1\pmod{7}$$

and a square

$$n\equiv t^2\equiv 0,1,4,2\pmod{7}$$

Therefore $n$ is either $7k$ or $7k+1$

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You forgot "square" after "simultaneously....". Note that it can be proved without brute-force checking all remainders/classes - see my answer. –  Bill Dubuque Apr 29 at 14:58

Suppose that $\,\color{#c00}{a^{k}} =n = \color{#0a0}{b^{k+1}}$ and $\ k(k\!+\!1)+1 = p\,$ is prime. $ $ By $\ \color{#c0d}{\mu F} = $ little Fermat

mod $\,p\!:\ \,n\not\equiv 0\,\Rightarrow a,b\not\equiv 0\,\Rightarrow\ n\equiv\dfrac{n^{\large k+1}}{n^{\large k}\ \ \ \ }\equiv \dfrac{(\color{#c00}{a^{\large k}})^{\large k+1}}{(\color{#0a0}{b^{\large k+1}})^{\large k}}\equiv\dfrac{a^{\large p-1}}{b^{\large p-1}}\,\overset{\color{#c0f}{\mu F}}\equiv\, \dfrac{1}{1}\equiv 1\ $

Hence $\ n\equiv 0\ $ or $\,1\pmod p,\ $ i.e. $\ n = pj\,$ or $\,1\!+\!pj\,$ for some $\,j\in \Bbb Z.\ \ $ QED

Yours is the special case $\ k = 2,\,$ so $\, p = 7.$

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If an integer $a$ is simultaneously a square and a cube, it must be the sixth power of an integer $b$

Now as $7$ is prime either $7|b$ or $(7,b)=1$

If $7|b, 7|b^n$ for integer $n\ge1$

Else by Fermat's Little Theorem , $7|(b^{7-1}-1)$

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Check the integers mod $7$. Which are squares and which are cubes:

$$0^2=0^3=0$$ $$1^2=1^3=1$$ $$2^2 = 4,\qquad 2^3=8\equiv 1$$

$$3^2 = 9 \equiv 2,\quad3^3=27\equiv 6$$ $$4^2 = 16 \equiv 2,\quad 4^3= 64 \equiv 1$$ $$5^2 = 25 \equiv 4,\quad 5^3= 125 \equiv 6$$ $$6^2 = 36 \equiv 1,\quad 6^3=216\equiv 6.$$

So only $0, 1, 2$ and $4$ are squares, only $0, 1$ and $6$ are cubes (mod $7$). The ones that are both give the wanted answer.

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I do not find this to be a convincing reason why squares and cubes mod 7 all follow this pattern. –  VF1 Apr 30 at 4:08
    
If $a\equiv b$, then $a^n\equiv b^n$, you only need to check one representatives of the residue classes ($0,1,...,6$). –  ploosu2 Apr 30 at 9:26

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