Ordered Pair Solutions for a2 + 2ab + b2 = a + 4b | Non-Negative Integers

[The legend]

Homework Statement

Find all ordered pair (x,y) of non negative integers satisfying
a² + 2ab + b² = a + 4b

Homework Equations

none

The Attempt at a Solution

(a+b)² = a + 4b
or
a(a-1) + b(b-4) + 2ab = 0

obviously (0,0) satisfies the equation. What about the rest?
Please help!

 

[HallsofIvy]

Think of this as $a^2+ 2ab+ b^2- a- 4b= a^2+ (2b-1)a+ (b^2- 4b)= 0$ and treat it as a quadratic equation in a with parameter b.

By the quadratic formula,
$$a= \frac{1- 2b\pm\sqrt{(2b-1)^2- 4(b^2-4b)}}{2}= \frac{1- 2b\pm\sqrt{12b+ 1}}{2}$$.

In order that a be a non-negative integer, it certainly must be rational which means that 12b+ 1 must be a perfect square. Writing $12b+ 1= n^2$ we must have $12b= n^2- 1= (n- 1)(n+ 1)$. Look at values of n such that $n^2- 1$ is a multiple of 12.

 

[The legend]

oh! thanks ... i got the answer solving!
Thanks again

 

[Mentallic]

You can also try this:
If we consider the left side as being (a+b)² then we know that any quadratic grows faster than a linear equation, a+4b so we can try and find at which point the quadratic will beat the linear equation and thus test all solutions under that certain point.

Since on the right hand side, 4b is greater than a (for the same integer values) any values other than a=0 would make the quadratic grow faster than the linear equation's fastest possible growth, so we just need to consider a=0.

So, b²=4b, is what we have reduced it to.
Now we have to find where (b+1)²>4(b+1)

b²+2b+1>4b+4
b²-2b-3>0
(b+1)(b-3)>0
thus b>3

Which means that b=4 would be the highest possible value that we should check. Thus from this point on we just check the solutions (a,b) up to a+b=4.

Now I might have broken some rules here, because I was just going off my intuition so yeah please correct me if I'm wrong or have made a false assumption anywhere.

 

[The legend]

ok...
but wasn't the assumption made on a=0?

 

[Mentallic]

Not in that way, I was assuming a=0 to find the maximum possible value of the linear equation a+4b, compared to the quadratic (a+b)². For b=4 this is the maximum since using any other values for a in a+b=4 will yield a lower number in the linear equation a+4b.

Yeah.. just trying to get my thoughts into writing here...

 

[The legend]

Right!
Now i get it!
Thanks a lot!

 

[Mentallic]

No problem