# Problem 4132 from Crux Mathematicorum

### Solution 1

Case 1: $\mathbf{2a^3+b^2=a^2+b^2}$

In this case, $2a^3=a^2\,$ and, since $a\in\mathbb{Z},\,$ $a=0.\,$ It follows that $2a^3b^2+ab^2+2b^4=3b^4\,$ which is divisible by $b^2=a^2+b^2.$

Case 2: $\mathbf{2a^3+b^2\ne a^2+b^2}$

In this case, $2a^3+b^2)=k(a^2+b^2),\,$ where $k\in\mathbb{Z}\setminus\{1\},\,$ implying $\displaystyle b^2=\frac{a^2(2a-k)}{k-1},\,$ or $\displaystyle a^2+b^2=\frac{a^2(2a-1)}{k-1}.\,$ So, we have

\displaystyle\begin{align} 2a^3b^2+ab^2+3b^4 &= b^2(2a^3+a^3b^2)\\ &=\frac{a^2(2a-k)}{k-1}\left[2a^3+a+\frac{3a^2(2a-k)}{k-1}\right]\\ &=\frac{a^3(2a-k)}{k-1}\cdot\frac{2(k+2)a^2-3ak+k-1}{k-1}\\ &=\frac{a^3(2a-k)}{k-1}\cdot\frac{(2a-1)[(k+2)a-(k-1)]}{k-1}\\ &=\frac{a^2(2a-1)}{k-1}\cdot\frac{(2a-k)[(k+2)a^2-(k-1)a]}{k-1}\\ &=(a^2+b^2)\cdot\frac{(2a-k)[(k+2)a^2-(k-1)a]}{k-1}. \end{align}

### Solution 2

Since $b^2-2ab^2=(2a^3+b^2)-2a(a^2+b^2),\,$ it follows from $(a^2+b^2)|(2a^3+b^2)\,$ that

(1)

$(a^2+b^2)|b^2(1-2a).$

Further, $ab^2+2b^4=ab^2(1-2a)+2b^2(a^2+b^2),\,$ thus, by (1),

(2)

$(a^2+b^2)|(ab^2+2b^4).$

Finally,

$2a^3b^2+ab^2+3b^4 = (ab^2+2b^4)+b^2(2a^3+b^2)$

so that $(a^2+b^2)|(2a^3b^2+ab^2+3b^4)\,$ by (2) and by the hypothesis.

### Acknowledgment

The above problem, with a solution (Solution 1), has been kindly communicated to me by Leo Giugiuc. Solution 2 is by Lorenzo Villa.