An Equation with Integer Solutions? (b^2-4ac=23)

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#NumberTheoryProblems #DiophantineEquations

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[littlefermat]

Cool problem 😎
Useful technique: it is always good to take mod the coefficients of an equation.
In our problem by simply taking mod 4 we get b^2=3 (mod 4), but 3 is not a quadratic residue for 4. Hence no solution.
And so we are done!

[ahmadkalaoun]

Alternative solution:
b²-4ac=23⇒
b² is congruent to 3 (mod 4 )
And you can easily check there is no such integers

[alnitaka]

We can go even further. If b^2-4ac = D, then D = 0 or 1 mod 4, because mod 4 it is b^2=D, implying that D is a square mod 4, so it is congruent to 0 or 1 mod 4, because those are the only squares mod 4.

[engjayah]

@3:22 further simplifying we get 2n(n+1) = 2ac + 11
Now LHS is even and RHS is odd; hence, by contradiction no solutions.

[gdtargetvn]

Since b^2 is odd, b is also odd.
Let b=2k+1 (k is an integer)
We now have
(2k+1)^2-4ac=23
-> 4k^2+4k+1-4ac=23
-> k^2+k-ac=11/2, which has no solution since k, a and c are integers

[bonwood]

Set 𝜶, 𝜷 are the solutions of the original equation. Divided by a^2 on both side of b^2 - 4ac =23 then we get b^2/a^2 - 4 c/a = 23/a^2. According to Vieta's formulas, 𝜶+𝜷= - b/a, 𝜶𝜷= c/a. Therefore (𝜶+𝜷 )^2 - 4𝜶𝜷 = 23/a^2, i.e. (𝜶-𝜷)^2 = 23/a^2. a is an integer and 23 is a prime. Therefore 𝜶, 𝜷 are not integers.

[quyenhoang]

Divided by 2 we have left side is even and right side is 11,

[honortruth]

The left side must be odd and b must be at least 7. Let b = 2n - 1, then b^2 = 4n^2 - 4n + 1 = 23. Subtract 1 from both sides and factor 4 on the left. 4(n^2 - n) = 22 —> has no solutions over the integers.

[pierregrangier]

One of the easiest exercices on your channel

[pratimakumari]

Which app do you use? Because I want to create a new channel of simple riddles of mathematics and I need these type of app it is available on android?

[ben]

How about if it were 24 instead of 23 ? I am just curious

[hemarajue]

I think b^2-4ac=5 has integer solutions. Since 5 can be expressed as 3^2-4(1)(1)=5.

[manojsurya]

Wow man odd n even made it simple, I thought u wud do some system of equations.🤩

[Grundini]

Since b^2-4ac is the discriminant of the Quadratic equation wouldn't it then have to be a perfect square for the equation to come out with integer solutions?

[rakenzarnsworld]

b must be odd number
when b=1, 4ac=-22
No integer solutions for a, c
when b=3, 4ac=-14
when b=5, 4ac=2
when b=7, 4ac=26
when b=9, 4ac=58
when b=11, 4ac=98
.
.
.
when b=n, 4ac=-23+n^2
when n is a odd number, n^2 must be 4k+1 (k is an integer)
3^2=1^2+8
5^2=3^2+16
7^2=5^2+24
n^2=(n-2)^2+8(1/2n-1/2)
THERE ARE NO SOLUTIONS!

[yoav]

This is easy but not so nice because there are no solutions😔 but you are nice so it is ok🙃😃

[giuseppemalaguti]

Non è ppssibile perché (b^2-23)/4 non può essere un prodotto tra imteri.... Infatti se b è pari b^2/4 e un ontero-23/4 non può essere intero.... Se b è dispari b^2/4 e 25 mentre 23/4 è 5, 75 quindi la differenza non è un intero e quindi non può essere um prodotto di imteri, come nel precedente caso

[zarifzawad]

Please upload how to solve ax^3+bx^2+cx+d=0

[Grassmpl]

3, equivalently 23, is not a quadratic residue mod 4. Thus, no solution exists.

[МаксимАндреев-щб]

b^2=0 or 1 (mod 4)
then b^2-4ac=0 or 1 (mod 4)
but 23=3(mod 4)
There are no integer solutions