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Statement

For positive integers $a,b$, the integers $a^2+ab+b^2$ and $a(a+b)$ cannot both be squares.

Source and scope. Beeson–Laczkovich–Zhang, arXiv:2604.03609v3, Proposition 19, p. 10. Two complete reductions are given below, with the classical Diophantine theorem and the elliptic-curve group data explicitly treated as external inputs. Their proofs are not reproduced.

Reduction to a classical Diophantine equation

Divide $a,b$ by their greatest common divisor. Both square conditions are unchanged, so assume $\gcd(a,b)=1$. If $a(a+b)$ is square, its coprime factors are squares: $a=m^2$, $a+b=n^2$, with $n>m>0$. Substitution in $a^2+ab+b^2=c^2$ gives

$$c^2=n^4-m^2n^2+m^4.$$

The external result cited in the source is that the positive integer solutions to this equation have $m=n$ (L. E. Dickson, History of the Theory of Numbers, vol. II, p. 638). That contradicts $n>m$.

Elliptic-curve alternative

Set $t=n/m>1$ and $s=c/m^2$. Then $s^2=t^4-t^2+1$. By Lemma 18, $(x,y)=(2t^2-2s-1,2tx)$ lies on

$$E:\quad y^2=x^3+2x^2-3x.$$

The external rank and torsion data are $\operatorname{rank}E(\mathbb Q)=0$ and $\#E(\mathbb Q)_{\rm tors}=8$. They are recorded in the source and in LMFDB 24.a4. Its model $y^2=X^3-X^2-4X+4$ is obtained by $X=x+1$. Thus all rational points are torsion. The eight distinct points

$$\mathcal O,\ (0,0),\ (1,0),\ (-3,0),\ (-1,\pm2),\ (3,\pm6)$$

satisfy the equation and therefore exhaust $E(\mathbb Q)$. For $x\ne0$, the inverse $t=y/(2x)$ gives only $t=0,1,-1$. If $x=0$, then $s=t^2-1/2$, and comparison with the quartic would give $1/4=1$, impossible. The image of a finite rational $(t,s)$ is affine, so it cannot be $\mathcal O$. None of the possible values of $t$ is greater than $1$. This gives the same contradiction by a different method.

The paper refers to a descent procedure in Silverman–Tate, pp. 91–94, and to the Nagell–Lutz theorem on p. 56 of that book, but does not print that descent calculation. The rank/torsion values are imported here from the identified curve record; no independent computer calculation of them is claimed.

Bears on. Problem 633.