A Fun Diophantine Equation (Not From Harvard 😜)
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a^2+2ab+b=44
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A slightly different method.
a^2 + 2ab + b = 44
a^2 + b(2a+1) = 44
// 2a+1 is a factor of 4a^2-1 (difference of two squares)
// multiply by 4 and subtract 1
(4a^2-1) + 4b(2a+1) = 175
// extract common factor
(2a+1)(2a + 4b - 1) = 175
// sum of factors = 4a + 4b
// let the factors be p & q
// find a+b directly
a+b = (p+q)/4
The rest follows...
a^2 + 2ab + b = 44
(a + b)^2 = b^2 - b + 44
(a + b)^2 = (b - 1/2)^2 - 1/4 + 176/4
(a + b)^2 = (b - 1/2)^2 + 175/4
4(a + b)^2 = (2b - 1)^2 + 175
(2a + 2b)^2 - (2b - 1)^2 = 175
(2a + 1)(2a + 4b - 1) = 5*5*7
a + b = [±44; ±10; ±8]
a little more explanation
let c=2a+1
d=2a+4b-1
c+d=4a+4b
(c+d)/4 = a+b
(c,d) = (1,25×7) or (5,5×7) or (7,25)
so c+d= 176 or 40 or 32
(c+d)/4= +-44 or +-10 or +-8
(c+d)/4=a+b
Answer
a+b=+-44 or +-10 or +-8
check
a=(c-1)/2
c=+-1, +-5, +-7, +-25, +-35, +- 175
a=0;-1, 2;-3, 3;-4, 12,-13, 17;-18, 87;-88
b=44;-43,
@@davidseed2939 Nicely done. I did it the hard way and got the same answers from 12 factorisations.
@davidseed2939 c,d can be 12 different pairs. I think you have them all covered but it looks like just 3 in one line.
That's so cool!
Another commenter worked out (as I did) that you can express b as a function of a thus: b = (44 - a*a)/(2a + 1)... Hence you can work out that a + b = (44 + a*a + a)/(2a + 1). The integer solutions with (a,b) equal to either (2,8) or (3,5) satisfy this formula for a+b .
The following method is easy to remember and works best if a and b are restricted, esp if a, b, are integers >/= 0...
Isolate b so that b = (44 - a²)/(1+2a)
Set >/= 0 and solve the inequality so that a < 6
Now solve for a = 6, 5, 4, 3, 2, 1 . . . making a table is the best way
I got (3, 5) and (2, 8)
3 works. a = 3 and b = 5, for example.
I plugged in 4 for a and 8 for b, so a+b=4+8=12.
Are you crazy 🤣 how can you put a as 4 and b as 8 it is not satisfying
a=2@@MaheshKumar-lx1ku
a = 0, b=44
a = 2, b = 8
a = 3, b= 5
I am sure there are others.
a + b = 44
a + b = 10
a + b = 8
Logarithmic inequalities, please.
If only the first equation had b^2 instead b.
I thought it was until I read your message!
😂
a+b=5+4=9 or
a+b=2+8=10.......May be
Explain later...
It is surprising how many really bad maths videos there are on YT.
Here is a systematic approach that is quick short and easy.
See that a must be less than 7.
Try a=6 and you get 13b = 8. No good
Try a = 5 and you get 11b = 19. No good
Try a = 4 and you get 9b = 28. No good
Try a=3 and you get 7b = 35 so b = 5.
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Do you think your solution is a good one? 🤔