How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$

Question

Given $27w^2+20w+35 \equiv 0 \pmod{23}$

How to solve this quadratic congruence? All hints are welcome. I completed the square ,so I obtained $27(x+\frac{10}{27})^2+\frac{845}{27} \equiv 0 \pmod{23} $ I need to get 2 solutions in the least residue system which are the correct answers: $w\equiv 6 \pmod{23}$ and $w \equiv 12 \pmod{23}$

ALWAYS reduce to smaller numbers. That's always a gimmee. $27w^2 +20w +35 \equiv 4w^2 -3w + 12$ or $4w^2-3w -11$. And as a hint the quadratic formula and factoring are still valid. — fleablood, Nov 27 '20 at 19:04

score 4 · Accepted Answer · answered Nov 27 '20 at 18:22

4

For ease of manual computation we rewrite the equation as $$4w^2-3w+12\equiv0\bmod23$$ Divide by the leading coefficient, i.e. multiply by $4^{-1}=6$: $$w^2+5w+3\equiv0\bmod23$$ Now apply the quadratic formula: $$w\equiv\frac{-5\pm\sqrt{13}}2\bmod23$$ We need to work out the square roots of $13$ in $\mathbb Z_{23}$. $6$ is easily verified as one root, so $-6$ is the other: $$w\equiv\frac{-5\pm6}2\equiv9\pm3\bmod23$$

answered Nov 27 '20 at 18:22

Parcly Taxel

103,344

thank you, it is right too – Charlie Van Basten Øydne Nov 27 '20 at 18:36
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How do we justify that $\pmod {23}$ that values have only $2$ square roots? – fleablood Nov 27 '20 at 19:03
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@fleablood Because $23$ is prime. If we have $x,y, x^2 \equiv y^2 \equiv 13$ then $23|(x^2-y^2)=(x+y)(x-y)$ so either $x\equiv y$ or $x\equiv -y$. There cannot be a third root. – Neat Math Nov 27 '20 at 19:23
That should be part of your explanation. – fleablood Nov 28 '20 at 01:28

Neat Math · Answer 2 · 2020-11-27T19:11:16.190

3

Hint:

$$\pmod{23}: 4w^2-3w+12\equiv 0 \implies 8w^2-6w+1\equiv 0 \implies (2w-1)(4w-1)\equiv 0. $$

Update To justify why I multiply 2 to $4w^2-3w+12$, it's easier to work with whole numbers than fractions, so to complete square while keeping every coefficients integer we multiply by 16:

$$16(4w^2-3w+12)=64w^2-48w+192=(8w-3)^2+183\equiv (8w-3)^2-1 = (8w-2)(8w-4)=8(4w-1)(2w-1) \pmod{23}$$

and now you see why.

Update 2: I like Parcly Taxel's way of making the quadratic monic first:

$$w^2+5w+3\equiv0\pmod{23}$$

After that it can be done a little faster:

$$w^2-18w+3\equiv 0 \implies (w-9)^2 = 78\equiv 9 =3^2 \implies (w-6)(w-12) \equiv 0 \pmod{23}$$

edited Nov 27 '20 at 19:11

answered Nov 27 '20 at 18:23

Neat Math

4,790

thank you, it is right – Charlie Van Basten Øydne Nov 27 '20 at 18:29
1

Updated. I didn't multiply by 2 from nowhere. – Neat Math Nov 27 '20 at 18:52
I got it. thank you – Charlie Van Basten Øydne Nov 27 '20 at 18:59
I took $u=8w-3$ therefore $u^2 \equiv -183 mod 23$ and notice that $23-1=22$ so $ u \equiv 1 \pmod{23}$ and $u \equiv 22 \pmod{23}$ and get back to substitution: $ 8w-3 \equiv 1 \pmod{23}$ which is $ w \equiv 12 \pmod{23}$ and $ 8w-3 \equiv 22 \pmod{23}$ which is $ w \equiv 6 \pmod{23}$ and it´s done. – Charlie Van Basten Øydne Nov 27 '20 at 21:15
Please check my second update. A combination of my method with that of another answerer which gives you the fastest solution. – Neat Math Nov 27 '20 at 21:22

md2perpe · Answer 3 · 2020-11-27T19:39:57.130

1

Since $27 \equiv 4$ we can write the equation as $4w^2 + 20w + 35 \equiv 0.$ Completing the square gives $(2w+5)^2 + 10 \equiv 0,$ i.e. $(2w+5)^2 \equiv -10.$ But $-10 \equiv -10+2\cdot 23=36=6^2,$ so $2w+5\equiv\pm 6,$ i.e. $2w=-5\pm 6.$

Case $+$: $2w=-5+6=1\equiv 1+23=24=2\cdot12$ so $w\equiv12.$

Case $-$: $2w=-5-6=-11\equiv -11+23=12=2\cdot6$ so $w\equiv6.$

Thus the solutions are $w=12$ and $w=6$.

edited Nov 27 '20 at 19:39

answered Nov 27 '20 at 19:05

md2perpe

26,770

thank you, it is interesting how many ways to solve it – Charlie Van Basten Øydne Nov 27 '20 at 19:31

How to solve this quadratic congruence? $27w^2+20w+35 \equiv 0 \pmod{23}$

3 Answers3