This is all true but who would think of that method? Obvious approach is to solve for b: 2b=(44-a^2)/(a+1/2) Then split into partial fractions. 2b=-a + 1/2 + 175/(a+1/2) This describes a hyperbola which crosses the a axis at √44 so there are just 6 integer cases to try, and the two correct ones turn up immediately
Good problem. You can represent as u² - v² = 175 where u = ± 2(a+b) and v = ± (2b-1) If the two factors are α & β, αβ = 175, then examine u = (α + β)/2 and v = (α - β)/2; u and v must be even and odd respectively; a & b must be > 0 This narrows down to 5•35 and 7•25; (a,b) = (2,8), (3,5) I think the full set of integer solutions is (a,b) = (0,44),(-1,-43),(2,8),(-18,8),(17,-7),(-3,-7),(12,-4),(-4,-4),(3,5),(-13,5)
Excellent solution, I like how you tested all the possible values of b! 💯🙏Thanks for the detailed explanation, I appreciate the step-by-step breakdown! 👍
let's take the edge cases off the list: a=0 =>b=44 Otherwise b=(44-a^2)/(2a+1)=(-a^2-a/2+a/2+1/4-1/4+44)/(2a+1)=-a/2+1/4+175/(8a+4)=(1-2a)/4+175/(8a+4)=(1-2a)/4+(7*25)/(8a+4) We need at least 2a+1 to divide 7*25 but also the fractions to add to a whole 2a+1 belongs to set{-175;-35;-25;-7;-5;-1;1;5;7;25;35;175} => a belongs to {-88;-18;-13;-4;-3;-1;0;2;3;12;17;87} Values of b corresponding to a from set: {44+1/4-1/4; 1/4+9-5/4; 27/4-7/4; 9/4-25/4;7/4-35/4;3/4-175/4;1/4+175/4;-3/4+35/4; -5/4+25/4;-23/4+7/4;-33/4+5/4;-173/4+1/4} B belongs to {44,8,5,-4;-7;-43;44;8;5;-4;-7;-43} => (a,b)€ { (-88,44),(-18,8),(-13,5),(-4,-4),(-3,-7),(-1,-43),(0,44),(2,8),(3,5),)(12,-4),(17,-7),(87,-43)}
I see I could have got to your form by multiplying by 4 and moving the first fraction to one side then multiplying by denominator. So your method works too if you can accept that 2a+4b-1 can be larger than 2a+1 and the factors may be negative I also gave a fleeting thought to using quadratic formula, but here might be a good place for it: A1,2=(2b+-√(4b^2-4(b-44))/2=b+-√(b^2-b+44) this is not particularly helpful because the discriminant is always >0 and we get no good limitations for the number of values to check
This is all true but who would think of that method? Obvious approach is to solve for b:
2b=(44-a^2)/(a+1/2)
Then split into partial fractions.
2b=-a + 1/2 + 175/(a+1/2)
This describes a hyperbola which crosses the a axis at √44 so there are just 6 integer cases to try, and the two correct ones turn up immediately
Good problem. You can represent as u² - v² = 175 where u = ± 2(a+b) and v = ± (2b-1)
If the two factors are α & β, αβ = 175, then examine u = (α + β)/2 and v = (α - β)/2; u and v must be even and odd respectively; a & b must be > 0
This narrows down to 5•35 and 7•25; (a,b) = (2,8), (3,5)
I think the full set of integer solutions is (a,b) = (0,44),(-1,-43),(2,8),(-18,8),(17,-7),(-3,-7),(12,-4),(-4,-4),(3,5),(-13,5)
Nice solution 😍🤩🥰✅😎
As a>=1, the original equality implies that b < 44/3, i.e., b
Excellent solution, I like how you tested all the possible values of b! 💯🙏Thanks for the detailed explanation, I appreciate the step-by-step breakdown! 👍
There is infinite number of solutions. Aa an example, try plug x=2, you get y=8.
Great observation! It's a good idea to check for different solutions. 😊
(20)+2ab+(22)=(10^10)+2ab+(11^11) (5^5^5^5)+2ab×(5^6^5^6) (1^1^1^1)+2ab+(1^3^2^1^3^2) 1ab+(1^13^2) (3^2) (ab ➖ 3ab+2).
let's take the edge cases off the list:
a=0 =>b=44
Otherwise
b=(44-a^2)/(2a+1)=(-a^2-a/2+a/2+1/4-1/4+44)/(2a+1)=-a/2+1/4+175/(8a+4)=(1-2a)/4+175/(8a+4)=(1-2a)/4+(7*25)/(8a+4)
We need at least 2a+1 to divide 7*25 but also the fractions to add to a whole
2a+1 belongs to set{-175;-35;-25;-7;-5;-1;1;5;7;25;35;175}
=> a belongs to {-88;-18;-13;-4;-3;-1;0;2;3;12;17;87}
Values of b corresponding to a from set:
{44+1/4-1/4; 1/4+9-5/4; 27/4-7/4; 9/4-25/4;7/4-35/4;3/4-175/4;1/4+175/4;-3/4+35/4; -5/4+25/4;-23/4+7/4;-33/4+5/4;-173/4+1/4}
B belongs to {44,8,5,-4;-7;-43;44;8;5;-4;-7;-43}
=> (a,b)€ { (-88,44),(-18,8),(-13,5),(-4,-4),(-3,-7),(-1,-43),(0,44),(2,8),(3,5),)(12,-4),(17,-7),(87,-43)}
Awesome producing all integer solutions 🙏🤩🤩🙏👏
I see I could have got to your form by multiplying by 4 and moving the first fraction to one side then multiplying by denominator.
So your method works too if you can accept that 2a+4b-1 can be larger than 2a+1 and the factors may be negative
I also gave a fleeting thought to using quadratic formula, but here might be a good place for it:
A1,2=(2b+-√(4b^2-4(b-44))/2=b+-√(b^2-b+44) this is not particularly helpful because the discriminant is always >0 and we get no good limitations for the number of values to check
x = k
y = (44 - k²)/(2k + 1)
k € R
k is not equal (- 1/2)
I got a=-4 and b=-4 as I took b-44=b^2. I am not confident with the accuracy of my method.
Then I used (a+b)^2 identity which equals zero after we take 44 to LHS.
And solved it as a quadratic equation.
You're welcome to show your full workings so that I can check on your accuracy
B=/P !!!
Irene, good evening. I'm on the process of learning correct pronunciation. Thanks for your concern and support 💕😍🤩👏🙏
Dieses Ratespiel hat nichts mit Mathematik zu tun!
It's not guess work but arithmetic principles✅