0
$\begingroup$

A natural number is a perfect square as well as a perfect cube. Show that it is $0$ or $1$ $($mod $7$$)$.

I tried the following.
There are integers numbers $x,y$ such that $n=x^{2}=y^{3}.$ By using Euclidean division, then $x$ and $y$ can be written as $7k,7k+1,7k+2,7k+3,7k+4,7k+5$ or $7k+6$. I am trying some contradictory stuff.

I don't know from where to go from here. Please suggest some hints.

$\endgroup$
2
$\begingroup$

You know that all the squares are equals to $0,1,2$ or $4$ mod $7$ because:

$$0^2=0\pmod 7$$ $$1^2=1\pmod 7$$ $$2^2=4\pmod 7$$ $$3^2=2\pmod 7$$ $$4^2=2\pmod 7$$ $$5^2=4\pmod 7$$ $$6^2=1\pmod 7.$$

And all the cubes are equals to $0,1$ or $6$ mod $7$ because:

$$0^3=0\pmod 7$$ $$1^3=1\pmod 7$$ $$2^3=1\pmod 7$$ $$3^3=6\pmod 7$$ $$4^3=1\pmod 7$$ $$5^3=6\pmod 7$$ $$6^3=6\pmod 7.$$

So if your number is a square and a cube at the same time, then it is necessarily equal to $0$ or $1$ mod $7$.

$\endgroup$
  • $\begingroup$ I am embarrassed at missing this. Thanks for the simplistic answer. $\endgroup$ – Shraddheya Shendre Dec 23 '16 at 9:43
-1
$\begingroup$

So, $n$ has to be of the form $$z^{\text{lcm}(2,3)}$$ where $z$ is any integer

If $7|z$, then $z^6\equiv0\pmod7$

Else $7\nmid z\implies(7,z)=1$ using Fermat's little Theorem $z^{7-1}\equiv1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.