Prove that there are infinitely many triples $(,,)$ of integers such that $a^2+b^2+c^2=a^3+b^3+c^3$

Question

I have already made the observation that since the domain is real numbers, $a^2+b^2+c^2$ the three terms is always positive but $a^3+b^3+c^3$ since its to an odd power can have negative terms implying that at least one of $a$, $b$ or $c$ must be negative.

Answer 1

Looks like $c= -b$ is good choice. Expression becomes: \begin{eqnarray} a^2 + 2b^2 &= &a^3,\\ 2b^2& = &a^3 - a^2 \end{eqnarray}

Letting $b=ak$ gives $2k^2 = a-1$, which implies $$(a,b,c)=(2k^2 +1, 2k^3 + k, -2k^3 -k)$$

satisfies the condition for any integer $k$.

Answer 2

I generated lists of some integer triples that satisfy the equation and noticed what seemed to be two families of related triples. It is easy to verify that the two families of integer triples

$$ (a,b,c) = (-n(2n^2+1),\, (2n^2+1),\, n(2n^2+1)), \tag{1} $$ $$ (a,b,c) = (1-n(n^2-1)/2,\, -(n^2-1),\, 1+n(n^2-1)/2), \tag{2} $$

all satisfy the equation

$$ a^2+b^2+c^2=a^3+b^3+c^3 \tag{3} $$

for any integer $\,n\,$ and therefore there are an infinite number of integer triple solutions to the equation, but the solutions for $\,n<0\,$ give the same unordered triples as for $\,n>0.\,$

Note, for the first family, $\,a+c=0\,$ and for the second family $\,a+c=2$.

Aside from the obvious solution triples $$ (0,0,0),\,(0,0,1),\,(0,1,1),\,(1,1,1), $$ there are a few other solution triples such as $$ (-1,-1,2),\, (-95,76,76),\, (-98,-21,99),\, (-330,-275,385),\, (-332,123,327), \dots $$ which may have parametrizations. In fact,as Will Jagy comments, there is an elliptic curve which gives solutions to equation $(3)$ but an elliptic curve can have only a finite number of integer points. The curve is $$ E: y^2+y = x^3+x^2 \tag{4} $$ with generating point $\,P=(0,0).\,$ Verify that if $\,(x,y)\,$ satisfies equation $(4)$ then $\,(y+1,-x,-y)\,$ satisfies equation $(3)$. The point $\,3P=(1,-2)\,$ gives the solution $\,(-1,-1,2)\,$ and the point $\,7P=(21,-99)\,$ gives the solution $\,(-98,-21,99).\,$

Answer 3

$$a^2+b^2+c^2=a^3+b^3+c^3\tag{1}$$

Let $b=d-an, c=e+an$ then we get

$-a^3+(2n^2-3dn^2+1-3en^2)a^2+(-2dn+2en+3d^2n-3e^2n)a+e^2+d^2-d^3-e^3=0.$

Finding the integer soluions of simultaneous equation {$-2dn+2en+3d^2n-3e^2n=0, e^2+d^2-d^3-e^3=0$},
we get the only integer soluions $(d,e)=(0,0),(1,1).$

$\bullet\ (d,e)=(0,0)$

Let $b=0-an, c=0+an$ then we get $a = 1+2n^2.$
Hence $(a,b,c)=(1+2n^2, -(1+2n^2)n, (1+2n^2)n).$
This is a sansae's solution.

$\bullet\ (d,e)=(1,1)$

Let $b=1-an, c=1+an$ then we get $a = -4n^2+1.$
Hence $(a,b,c)=(-4n^2+1, 1-(-4n^2+1)n, 1+(-4n^2+1)n).$
This is a Somos's solution.