4
$\begingroup$

Prove the number $a = 2 \cdot 10^{2016} + 16$ cannot be a perfect square.

I tried studying $a \mod b$ for different integers $b$, without succes.

  • 1
    If $a=k^2$, then $(k-4)(k+4)=2\cdot 10^{2016}$, and there shouldn't be many factors of $2\cdot 10^{2016}$ with a difference of 8. – 2017-01-26
  • 0
    @Sawarnik You should convert that to an answer because the difference of $8$ means that only one of the factors can be divisible by $5$ ... – 2017-01-26
  • 0
    @MarkBennet I wrote an answer. Can you check it? :) – 2017-01-26

4 Answers 4

10

Note that we have $$ 2\cdot 10^{2016} + 16 \equiv 2\cdot(-1)^{2016} + 5 = 7\pmod {11} $$ but the only squares modulo $11$ are $0, 1, 3, 4, 5, 9$.

  • 1
    ${\rm mod}\ 11\!:\ a^{\large 2}\equiv 7\equiv -2^{\large 2}\,\overset{\,(\ \ )^{{\LARGE 5}}}\Longrightarrow\,1\equiv a^{\large 10}\equiv -2^{\large 10}\equiv -1\,\Rightarrow\!\Leftarrow,\ $ i.e. by [Euler's Criterion](https://en.wikipedia.org/wiki/Euler's_criterion) – 2017-01-26
  • 0
    It looks like i missed 11 :( – 2017-01-26
  • 0
    @EugenCovaci I added an answer giving a generalization (which sheds light on the above). – 2017-01-26
2

Hint $\ $ To prove $\,2\,(10^{\large 2N}\!+8)\,$ nonsquare it suffices to find a prime $p$ where it is nonsquare mod $p,\,$ e.g. if $\,2\,$ is nonsquare and $\,10^{\large 2N}\!\!+8$ is square (e.g. if $\,10^{\large 2N}\!\equiv 1),\,$ e.g. $\,p = 11\,$ (Arthur's answer).

1

Let $$k^2 = 2\cdot 10^{2016} + 16$$ Then we have $(k-4)(k+4)=2\cdot 10^{2016}$. Thus we have two factors of $2\cdot 10^{2016}$ with a difference of $8$.

Now if these factors, say $a$ and $b$, both contain $5$ in their prime factorization, then we have that $|a-b|$ is a multiple of $5$ which is not $8$. Thus let $a=2^k$ and $b= 2^m 5^n$, where $k,m,n$ are non-negative integers. We know that $ab = 2^{m+k}5^{n}= 2^{2017}5^{2016}$. Thus $m+k=2017$ and $n=2016$. Now:

$$a-b = 2^k - 2^m5^n = \pm 8$$ $$2^{k-3} - 2^{m-3}5^n = \pm 1$$

Its easy to see that $k\geq 3$ and $n\geq 3$, since both can't be smaller than $3$. Since $\pm 1$ is not divisible by $2$ we have that either $m=3$ or $k=3$.

Thus our possible solutions for $(m,k,n)$ is $(3,2014,2016)$ and $(2014,3,2016)$, neither of which satisfies the last equation. Hence, we have a contradiction.

  • 0
    I would just say the factors have a difference of $8$. Any common factor divides the difference. Therefore all the factors of $5$ must belong to the same factor, which will be divisible by $5^{2016}\gt 2^{4032}\gt 2^{2017}+8$ and the largest possible second factor is $2^{2017}$. I have used $5\gt 2^2$. – 2017-01-26
-1

We have

$$\sqrt2\cdot10^{1008}\le\sqrt{2\cdot10^{2016}+16}=\sqrt2\cdot10^{1008}\sqrt{1+8\cdot10^{-2016}}<\sqrt2\cdot10^{1008}(1+4\cdot10^{-2016})$$ or

$$0\le\sqrt{2\cdot10^{2016}+16}-\sqrt2\cdot10^{1008}<\sqrt2\cdot4\cdot10^{-1008},$$ which is equivalent to

$$\{\sqrt2\cdot10^{1008}\}<\sqrt2\cdot4\cdot10^{-1008}.$$ This would imply that the decimals of $\sqrt2$ are all zeroes from the $1008^{th}$ to the $2015^{th}$.

So $2\cdot10^{2016}+16$ is probably not a square. :-)

  • 0
    Stupid downvote. – 2017-01-30