3.8 *Extended investigations concerning the non-existence of perfect 2-error-correcting codes on q symbols*

I refer to section 3.2 of my thesis, see here a reference to my thesis (in the literature list of the article).

3.8.1 *** THEOREM

A perfect 2-code does not exist if q=33

PROOF Assume there exists such a code with q=33. Then we have from 3.2.2, 3.2.4 and 3.2.5

3.8.2 *** x_{1}x_{2} = 2.3^{β1}11^{β2}

3.8.3 *** βi = k + αi - 2 and α1 α2 = 0

From 3.2.7h) it follows that β1 > 0, so from b) it follows that x_{1} and x_{2} are both divisible by 3 but not both by 11.

Since α1 α2 = 0 it follows from f) that if αi > 0 then

3.8.4 *** p_{i}^{αi} = 1 (mod 32), so αi = 8s with s > 0.

Hence, if we replace k-2 by u and set {x_{1},x_{2}} = {x,y}, then from h) we see there are only two possibilities:

3.8.5(A) *** x = 2 3^{β}11^{u}, y = 3^{γ} where β + γ = u + 8s (case A),
or

3.8.5(B) *** x = 3^{β}11^{u}, y = 2 3^{γ} where β + γ = u + 8s (case B)

(s, β, γ positive; β + γ at least 8).

We shall distinguish between cases (A) and (B).

case (A)

In this case we find by substitution in e)

3.8.6 *** 33(2.3^{β}11^{u} - 3^{γ})^{2} - 2(2.3^{β}11^{u} + 3^{γ}) = 27.

If γ ∈ {1,2,3} then by h) we have a contradiction to 3.8.6.

So γ is at least 4 and we see by considering 3.8.6 modulo 81 that β is at most 3.

Now if u=0 then by h) γ = β + 1, so γ=4 and β=3, which doesn't comply with 3.8.6 either.

So u > 0 and we find from 3.8.6

3.8.7 *** -2.3^{γ} = 27 (mod 11), so γ = 1+5t for some positive t.

Hence, considering 3.8.6 modulo 121, since 3^{1+5t} = 3 (mod121), we find

3.8.8 *** 9.33 - 4.3^{β}11^{u} - 6 = 27 (mod 121), so 8.33 = 4.3^{β}11^{u} (mod 121).

Since u > 0 this yields u=1 and

3.8.9 *** 6 = 3^{β} (mod 11), which is impossible.

So case (A) is not possible.

case (B)

In this case we find by substitution in e)

3.8.10 *** 33(3^{β}11^{u} - 2.3^{γ})^{2} - 2(3^{β}11^{u} + 2.3^{γ}) = 27.

Now if u=0 then by h) we find β = γ+1. Then γ can't be greater than 3 because 27 isn't divisible by 81. But in this subcase γ ∈ {1,2,3} doesn't comply with 3.8.10 either.

So u must be positive. Hence from 3.8.6 we deduce

3.8.11 *** 3^{γ} = 7 (mod 11), which is impossible.

So case (B) is not possible either.

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3.8.12 *** THEOREM

A perfect 2-code does not exist if q=38

PROOF Assume there exists such a code with q=38. Then from 3.2.7a) and b) we find

3.8.13 *** {x_{1},x_{2}} = {2^{a},2^{b}19^{c}}, where (from h)) c is positive and a is at least 4.

Substituting in e) yields

3.8.14 *** 19(2^{a} - 2^{b}19^{c})^{2} - (2^{a} + 2^{b}19^{c}) = 16.

Considering this last equality mod 16 we get

3.8.15 *** 3(2^{b}3^{c})^{2} = 2^{b}3^{c} (mod 16), so 2^{b}3^{c}=11(mod 16) so b=0 and c=3+4t.

Considering 3.8.14 modulo 19 we find

3.8.16 *** 2^{a} = 3(mod 19), so a = 13+18u.

Hence we find

3.8.17 *** {x_{1},x_{2}} = {2^{13+18u},19^{3+4t}}.

Now in view of d) we must have that 2^{13+18u}(mod 37) is either 1 or 2, so 13+18u must be either 0 or 1 modulo 36. This doesn't hold because 13+18u is either 13 or 31 modulo 36.

So q=38 is impossible.

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3.8.18 *** THEOREM

A perfect 2-code does not exist if q=39

PROOF Assume there exists such a code with q=39. Then we have from 3.2.2, 3.2.4 and 3.2.5

3.8.19 *** x_{1}x_{2} = 2.3^{β1}13^{β2}

3.8.20 *** βi = k + αi - 2 and α1 α2 = 0

From 3.2.7h) it follows that β1 > 0, so from b) it follows that x_{1} and x_{2} are both divisible by 3 but not both by 13.

Since α1 α2 = 0 it follows from f) and h) that α1 > 0 and

3.8.21 *** 3^{α1} = 1 (mod 38), so α1 = 18s with s > 0.

Hence, if we replace k-2 by u and set {x_{1},x_{2}} = {x,y}, then from h) we see there are only two possibilities:

3.8.22(A) *** x = 2 3^{β}13^{u}, y = 3^{γ} where β + γ = u + 18s (case A),
or

3.8.22(B) *** x = 3^{β}13^{u}, y = 2 3^{γ} where β + γ = u + 18s (case B)

(u, s, β, γ positive; β + γ at least 19).

We shall distinguish between cases (A) and (B).

case (A)

In this case we find by substitution in e)

3.8.23 *** 39(2.3^{β}13^{u} - 3^{γ})^{2} - 2(2.3^{β}13^{u} + 3^{γ}) = 33.

Considering this mod 9 we see β=1.

From 3.8.22(A) and 3.2.7d) we find γ = 18t and u = 1+18v for some positive u,v. So

3.8.24 *** 39(2.3.13^{1+18v} - 3^{18t})^{2} - 2(2.3.13^{1+18v} + 3^{18t}) = 33. So

3.8.25 *** 3(2^{18v} - 3^{18t})^{2} = 2^{18v} + 3^{18t} (mod 11).

3.8.26 *** One of 18v, 18t is 90p and the other 54+90q for some positive p,q.

Now in view of h), taking logarithms, we have the following possibilities: