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Let $88k + 41$ be an integer where $k \in \mathbb{Z}$. Is there any value for $k$ such that $88k+41$ is a perfect square? I tried to find that using C++ program but it wasn't successful. I don't know how to find such $k$ or prove that it's impossible.

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4 Answers 4

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$88k+41\equiv8\pmod{11}$

Now for any integer $a,$

$$a^2\equiv0,1,4, 9,5,3\pmod{11}\not\equiv8$$

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  • $\begingroup$ Can you explain why we have $a^2\equiv0,1,4, 9,5,3\pmod{11}$, please? $\endgroup$
    – S.H.W
    Dec 9, 2019 at 14:52
  • $\begingroup$ @SHW, AS $$a\equiv0,\pm1,\cdots\pm5\pmod{11}$$ $\endgroup$ Dec 9, 2019 at 15:10
  • $\begingroup$ Okay, thanks a lot. $\endgroup$
    – S.H.W
    Dec 9, 2019 at 16:00
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Mod $11$ and you see it's impossible.

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  • $\begingroup$ Why it should be quadratic residue when we take Mod 11? $\endgroup$
    – S.H.W
    Dec 9, 2019 at 14:00
  • $\begingroup$ If you want it to be a perfect square, then it is by definition a quadratic residue. $\endgroup$
    – WhatsUp
    Dec 9, 2019 at 14:01
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In case the other response is too advanced, look at the remainders when you divide square numbers by $11$.
The remainders of $1,4,9,...,121$ are $1,4,9,5,3,3,5,9,4,1,0$ Notice the pattern reads the same backwards and forwards. That pattern repeats for 144,... 484.
The remainders are 0,1,3,4,5 or 9. What is the remainder when you divide your number by 11?

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  • $\begingroup$ Thanks, how we can prove that in general? $\endgroup$
    – S.H.W
    Dec 9, 2019 at 14:54
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    $\begingroup$ Show that $(11a+b)^2$ leaves the same remainder as $b^2$ because their difference is a multiple of $11$ $\endgroup$
    – Empy2
    Dec 9, 2019 at 18:07
  • $\begingroup$ @S.H.W It's a bit easier to use litte Fermat - see my answer. $\endgroup$ Dec 14, 2019 at 5:34
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$\!\bmod 11\!:\ a^{\large 2}\! = 88k\!+\!41 \equiv -3\,\overset{(\ \ )^{\Large 5}}\Longrightarrow\, a^{\large 10} \equiv (-3)^{\large 5} \equiv (-3)(-2)^{\large 2} \equiv -1\,$ contra little Fermat

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