In GF($2^4$) usinfg primitive polynomial g(x) = $x^4 + x^3 +1$, 2+3 = 1 since 010 + 011 = 001,

2 + 2 = 0

Is there any encoding to get 2 + 3 = 5 and 2 + 2 = 4 in GF.

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    $\begingroup$ Not in a binary field. $\endgroup$ – CodesInChaos Dec 18 '17 at 10:11

As stated by CodesInChaos in comment, what's asked can't be done with the same $+$ operation for the binary field $\operatorname{GF}(2^4)$ and the desired $+$ operation for common integers. Proof: in $\operatorname{GF}(2^4)$, $a+a=b+b$ for any $a$ and $b$, and that does not hold for any useful subset of integers.

However, we can map the elements of $\operatorname{GF}(2^4)$ less zero (0000) to $\{0,1,2\dots,13,14\}=\mathbb Z_{15}$, with multiplication in $\operatorname{GF}(2^4)$ mapping to addition modulo $15$ in $\mathbb Z_{15}$ . We take any generator $g$ of $\operatorname{GF}(2^4)$, e.g. 0010; map 0001 to $0$, $g=$0010 to $1$, $g^2=$0100 to $2$, $g^3$=1000 to $3$, $g^4$=1001 to $4$, and more generally $g^n$ to $n$.

Now, the operation in $\mathbb Z_{15}$ defined by getting back to $\operatorname{GF}(2^4)$ per the inverse of the above mapping on both arguments, multiplying the results in $\operatorname{GF}(2^4)$ , and getting to $\mathbb Z_{15}$ again by applying said mapping, is simply addition modulo $15$, and thus works as asked for small non-negative integer values.

Analogy: we can perform addition in the set $\mathbb R$ (the reals) as $x+y=\log(e^x\times e^y)$.

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