a=4, b=2 As per question a+2ab+b=22 Multiply 2 on both the sides3 2a+4ab+2b=44 Add 1 on both the sides3 2a+4ab+2b+1=44+1=45 2a(1+2b)+1(1+2b)=45 (1+2b)(2a+1)=9×5 or 15×3 Let's take 9×5 So, (1+2b)(2a+1)=9×5 So, 1+2b=9 and2 2a+1=5 2b=9-1 and 2a=5_1 2b=8 and 2a =4 b=4 and a=2( also vice versa)

The key is the solutions are integers. Simply solve for a in terms of b.No fancy manipulation necessary. Always look for simple solutions first. In addition,thats applicable to all linear Diophantine equations.

Excellent, as usual. What I learn is that I need to learn to be more creative. 😊

(2a+1)(2b+1)=45 (a;b)={(1;7),(7;1),(2;4),(4;2)}

Eccellente soluzione!

Is it called integer Solution?

I solved in this way: a + 2ab + b = 22 -> b(2a +1) = 22 - a -> b = (22 - a) / (2a +1). b must be integer so I can compute rhis table a | 2a +1 | 22 -a -------------------------------- 1 3 21 x (22-a divide 2a +1) b=7 2 5 20 x b= 4 3 7 19 4 9 18 x the same solutions switched 5 11 17 6 13 16 7 15 15 x 8 17 14 9 19 13 10 21 12

@harrymatabal8448

16 күн бұрын

Good work from you and Mr Giulio. Thanks

2a + 1 >= 3 is not true if a (or b) = 0, you missed the 1, 45 pair and 45, 1 pair or (0, 22) and (22, 0) as long as your solution is only non-negative integers

How did you come up with tricks like multiplying both sides with 2 and then adding 1 to both sides? I mean, it seemed so random what you did and yet effective. Is there any signal of when to use such tricks and which tricks to use?

@neon1300

16 күн бұрын

Same question here.

@immersivalgames2696

16 күн бұрын

What I think he did is that he factored a+2ab and got a(1+2b) or a(2b+1) and since there is still one b left over he matched the factor by multiplying both sides by 2 to get the 2b and added 1 to finally match the factor so he can get 2a(2b+1) + 1(2b+1) = 45 And simplified it to (2b+1)(2a+1)=45. This can also be done to 2ab + b but for the video it was done as a + 2ab.

@Cagouille79

16 күн бұрын

a(1+2b)+b=22 a=(22-b)/(1+2b) For each value of b you have a value for a except if 1+2b=0 (if b=-1/2) So, number of solutions = Infinite. Isn't it ?

@SubhroMandal-r8k

15 күн бұрын

Bro! Only positive integers are asked.

@TheFrewah

10 күн бұрын

You could do a(1+2b) +b =22. And now you see you want the single b to be 2b+1 so you multiply by 2 and add 1 and then you can factorise.

At a quick glance if a = 4 and b = 2 then 4+16 + 2 = 22. This gives two solutions: a = 4 and b = 2. a = 2 and b = 4.

a+b = 22-2ab, so a+b is divisible by 2. Therefore, both are even or both odd. a+b positive, so 2ab < 22 and ab < 11. There are very few positive pairs of (a,b) to test that are both odd or both even that and satisfy ab

@user-nr6tt5he7g

16 күн бұрын

Continuing your reasoning. Let a≡a' and b≡b' (mod 4), where 0

This is too complicated,it is because a and b are positive integer,just try a equal to 1 to 6, then you can find all answers

It's big a deal👏🏻👏🏻👏🏻 I have a math olympiad question, can I send it?

@maryjay7833

14 күн бұрын

Please send it here, we will either help or learn something😊

a=2 and b=4...Just be common sense!

a = 1 and b = 7 ?

@Mofiac

15 күн бұрын

And 22, 0. His answers don't work in the original equation .

@quantumbuddha777

14 күн бұрын

​@@Mofiaczero is not a positive integer

(1,7), (7,1), (0, 22), (22,0) are the only solutions. Your answers don't work with the original equation.

@maryjay7833

14 күн бұрын

Sorry my friend , but you are wrong, a and b must be positive. He solved it nicely

There are more answers: a=22 b=0 and simetrical a=0 b=22 The solve idea is very nice and creative.

@johnlv12

16 күн бұрын

a and b must be positive

Was your fourth step necessary. Wasting our time

You have used totally unexplained steps such as multiplying by two and adding one to both sides. These steps seem picked out of the air with no logical reasoning. You do this a lot on your channel and I’ve commented as such in precious videos. It would be very useful for me if you could explain the reasons why you choose each step please.