Determining a number is perfect square

Question

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.

score 4 · Accepted Answer · answered Dec 09 '19 at 14:02

4

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

Now for any integer $a,$

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

answered Dec 09 '19 at 14:02

lab bhattacharjee

274,582

Can you explain why we have $a^2\equiv0,1,4, 9,5,3\pmod{11}$, please? – S.H.W Dec 09 '19 at 14:52
@SHW, AS $$a\equiv0,\pm1,\cdots\pm5\pmod{11}$$ – lab bhattacharjee Dec 09 '19 at 15:10
Okay, thanks a lot. – S.H.W Dec 09 '19 at 16:00

score 3 · Answer 2 · answered Dec 09 '19 at 13:58

3

Mod $11$ and you see it's impossible.

answered Dec 09 '19 at 13:58

WhatsUp

22,201
19
48

Why it should be quadratic residue when we take Mod 11? – S.H.W Dec 09 '19 at 14:00
If you want it to be a perfect square, then it is by definition a quadratic residue. – WhatsUp Dec 09 '19 at 14:01

score 3 · Answer 3 · answered Dec 09 '19 at 14:09

3

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?

answered Dec 09 '19 at 14:09

Empy2

50,853

Thanks, how we can prove that in general? – S.H.W Dec 09 '19 at 14:54
2

Show that $(11a+b)^2$ leaves the same remainder as $b^2$ because their difference is a multiple of $11$ – Empy2 Dec 09 '19 at 18:07
@S.H.W It's a bit easier to use litte Fermat - see my answer. – Bill Dubuque Dec 14 '19 at 05:34

score 1 · Answer 4 · answered Dec 14 '19 at 05:33

1

$\!\bmod 11\!:\ a^{\large 2}\! = 88k\!+\!41 \equiv -3\,\overset{(\ \ )^{\Large 5}}\Longrightarrow\, a^{\large 10} \equiv (-3)^{\large 5} \equiv (-3)(-2)^{\large 2} \equiv -1\,$ contra little Fermat

answered Dec 14 '19 at 05:33

Bill Dubuque

272,048

Above is a special case of Euler's criterion (modular square test), see here. – Bill Dubuque Mar 29 '23 at 19:06

Determining a number is perfect square

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