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
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$88k+41\equiv8\pmod{11}$
Now for any integer $a,$
$$a^2\equiv0,1,4, 9,5,3\pmod{11}\not\equiv8$$
answered Dec 9, 2019 at 14:02
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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.WDec 9, 2019 at 14:52
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$\begingroup$ @SHW, AS $$a\equiv0,\pm1,\cdots\pm5\pmod{11}$$ $\endgroup$ Dec 9, 2019 at 15:10
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Mod $11$ and you see it's impossible.
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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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2$\begingroup$ Show that $(11a+b)^2$ leaves the same remainder as $b^2$ because their difference is a multiple of $11$ $\endgroup$– Empy2Dec 9, 2019 at 18:07
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$\begingroup$ @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
answered Dec 14, 2019 at 5:33
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$\begingroup$ Above is a special case of Euler's criterion (modular square test), see here. $\endgroup$ Mar 29 at 19:06