# Adem relations of Steenrod algebra

Let $$\mathrm{Sq}^i$$ the Steenrod operations. These are the Adem relations:

when $$i<2j$$ $$\mathrm{Sq}^i \mathrm{Sq}^j=\sum_{k \in \mathbb{Z}}\binom{j-k-1}{i-2 k} \mathrm{Sq}^{i+j-k} \mathrm{Sq}^k,$$ with the understanding that $$\binom{n}{m}=0$$ if $$m>n$$ or $$m<0$$, and $$\mathrm{Sq}^k=0$$ if $$k<0$$ (so in practice $$k$$ ranges from $$\max (0, i-j+1)$$ to $$\left.\left\lfloor\frac{i}{2}\right\rfloor\right)$$.

In this link https://dec41.user.srcf.net/misc/adem I found this consequences: \begin{aligned} & \mathrm{Sq}^1 \mathrm{Sq}^1=0 \\ & \mathrm{Sq}^1 \mathrm{Sq}^2=\mathrm{Sq}^3 \\ & \mathrm{Sq}^1 \mathrm{Sq}^3=0 \\ & \mathrm{Sq}^2 \mathrm{Sq}^2=\mathrm{Sq}^3 \mathrm{Sq}^1 \\ & \mathrm{Sq}^1 \mathrm{Sq}^4=\mathrm{Sq}^5 \\ \end{aligned}

The problem is that if i try to compute $$\mathrm{Sq}^1\mathrm{Sq}^3$$ (that should be $$0$$ as suggested) this is what i get:
Since $$\max (0, i-j+1)=\max(0,1-3+1)=0$$ and since $$\left\lfloor\frac{i}{2}\right\rfloor = \left\lfloor\frac{1}{2}\right\rfloor=0$$ then $$k=0$$, so $$\mathrm{Sq}^1\mathrm{Sq}^3=\binom{3-0-1}{1-2*0} \mathrm{Sq}^{1+3-0} \mathrm{Sq}^0=2\mathrm{Sq}^{4}$$.
What is the problem? Why don't I get $$0$$?

I've got a similar problem trying to compute $$\mathrm{Sq}^1\mathrm{Sq}^4$$ (that should be $$\mathrm{Sq}^5$$ as suggested):
Since $$\max (0, i-j+1)=\max(0,1-4+1)=0$$ and since $$\left\lfloor\frac{i}{2}\right\rfloor = \left\lfloor\frac{1}{2}\right\rfloor=0$$ then $$k=0$$, so $$\mathrm{Sq}^1\mathrm{Sq}^4=\binom{4-0-1}{1-2*0} \mathrm{Sq}^{1+4-0} \mathrm{Sq}^0=3\mathrm{Sq}^{5}$$.
Why don't I get $$\mathrm{Sq}^{5}$$?

The Steenrod algebra is an algebra over $$\mathbb{F}_2$$, so in particular the coefficients are elements of $$\mathbb{F}_2$$. In other words, $$2 \mathrm{Sq}^4 = 0$$ and $$3 \mathrm{Sq}^5 = \mathrm{Sq}^5$$ and your problems disappear.