Let $q$ be a power of $2$. Recently, Tu and Zeng considered trinomials of the form $f(X)=X+aX^{(1/4)q^2(q-1)}+bX^{(3/4)q^2(q-1)}$, where $a,b\in\Bbb F_{q^2}^*$. They proved that $f$ is a permutation polynomial of $\Bbb F_{q^2}$ if $b=a^{2-q}$ and $X^3+X+a^{-1-q}$ has no root in $\Bbb F_q$. In this paper, we show that the above sufficient condition is also necessary.

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关于类型(1∕4,3∕4)的Tu-Zeng置换三项式

让 $q$ 是 $2$ 的幂。最近，Tu 和 Zeng 考虑了 $f(X)=X+aX^{(1/4)q^2(q-1)}+bX^{(3/4)q^2(q- 1)}$，其中 $a,b\in\Bbb F_{q^2}^*$。他们证明了 $f$ 是 $\Bbb F_{q^2}$ 的置换多项式，如果 $b=a^{2-q}$ 和 $X^3+X+a^{-1-q}$在 $\Bbb F_q$ 中没有根。在本文中，我们证明了上述充分条件也是必要的。