A Number Theory Thing

The equation can be written as $(2a+b)^2 \; = \; b^2(2b-3)$ so we require $2b-3 = (2n+1)^2$ to be an odd perfect square. Thus we need $b \; = \; 2(n^2 + n + 1) \hspace{2cm} a \; = \; nb$ and since $a,b \ge 1$ and $b <2017$ we deduce that $1 \le n \le 31$ . Thus there are $\boxed{31}$ solutions.

If we re-arrange the equation we get $2(a^2+b^2+ab)=b^3$ , so $b$ must be even. Let us make the substitution: $a=xm$ and $b=2m$ , where $m$ is integer. From the original equation we have a new one, $\frac{x}{2}\left(1+\frac{x}{2}\right) +1 =m$ . We see that for $m$ to be an integer, $x$ has to be an even integer (this follows from $\frac{x}{2}\left(1+\frac{x}{2}\right)=m-1=integer$ ), so $x=2n$ , $m=n^2+n+1$ and $b=2(n^2+n+1)$ . Starting from $n=1$ we have to go up to $n=32$ to get a $b>2017$ . There are 31 solutions.

Cheater's solution:

First we need to solve for $a$ . Since we are cheaters and don't actually know math, this is easiest to do using wolfram alpha: like so

As we can see, $a = {1\over{2}} (\sqrt{b^2 (2b-3)}-b)$ . But wait, what about the other solution? Well... that other solution will never give us a positive integer, so we can safely ignore it.

So, now that we know $a$ and we have a reasonable number of possible b values, we can write a program to find the b values that give us a.

The following code outputs $31$ .

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from decimal import Decimal, InvalidOperation

# Given b, compute a without any floating point errors. It is unlikely that
# floating point would cause a problem, but it is best practice to avoid it if
# you can.
def f(b):
  b = Decimal(b)
  try:
    return ((b**2 * (2*b-3)).sqrt() - b) / 2
  except InvalidOperation:
    # Probably tried to find the sqrt of a negative number.
    return None

answers = []
for b in xrange(1, 2017):
  a = f(b)
  # If a is a positive integer, add to list of answers.
  if a != None and a > 0 and a == int(a):
    answers.append((a, b))

print len(answers)

A direct and unrefined solution:

Expand: $a^2 + b^2 + a^2 + 2ab+ b^2 = b^3$

Gather and divide by 2: $a^2 + ab + (b^2 - \frac{b^3}{2}) =0$

Solve for $a$ : $a = \frac{ - b \pm \sqrt{ 2b^3-3b^2} }{2} = \frac{b}{2} (\pm \sqrt{2b-3} -1 )$

Since $a$ is positive, only + possible in $\pm$ .

Hence $\sqrt{2b-3}$ is an integer $>1$ (if it equals 1 then $a=0$ ).

$n = \sqrt{2b-3} \implies b = \frac{n^2+3}{2}$ .

Note that $n$ must be odd and $\geq 3$ . This means $b$ will always be even (a odd square +3 is always divisible by 4) and hence $b/2$ is always an integer. So these numbers (i.e. $\big( \frac{n^2+3}{4} (n+1), \frac{n^2+3}{2} \big)$ ) are exactly the integral positive solutions.

Look what is the largest $n$ : $\frac{n^2+3}{2} < 2017 \implies n \leq \sqrt{2 \cdot 2017 -3 } \approx 63.5$

There are 31 odd numbers between 3 and 63. So the answer is 31.

We have 2 (a^2 + b^2 +ab) = b^3. This means b has to be an even number. Let b be 2N. Solving the quadratic equation in a, we get a^2+2Na +(4N^2 -4N^3) = 0. This yields, a = -N + N sqrt(4N-3), taking only the positive values of a. Since a is a positive integer, 4N-3 has to be a perfect square. Let 4N-3 = K^2. Knowing that 0<N<=1008, we see that K can be only up to 63. Also every 2nd value of K from 1 to 63, yields integer solutions. Also K =1 is not valid as that yields a = 0 which is not positive. Hence we conclude that we have (63-1)/2 = 31 distinct solutions.

int main() { int a,b,x,y=0;

for(a=1;a<=201217;a++)

{

    for(b=1;b<2017;b++)

    {

        x=pow(a,2)+pow(b,2)+pow((a+b),2)-pow(b,3);

        if(x==0)
                     y++;
    }
}

cout<<"\n\n"<<y;

return 0;

}