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Well i personally believe that:

This procedure can be done for all n. >!Correct me if im wrong please.

Something we will need later:

T1: Two perfect squares differ by >!atleast 3. Proof: x >=1: (x+1)^2-x^2= >!2x+1 >= 3

Ok the base my proof is:

I will use induction on n. The >!induction base can be n=9 as in the >!question. In the induction step we >!know that the hypothesis is true for >!n=k-1 now we need to prove the >!hypothesis for n =k.

Ok now the proof begins:

Now we construct our 3 sets for >!n=k-1. Now we name the sets :A,B,C. >!Now if the numbers 2 and 1 are in A >!and we add k to A, and S(A) becomes >!a square according to T1 if we take 2 >!or 1 S(A) will be not a squre and if we >!add for exmaple 2 to another set and >!it doesnt become a square then its ok >!otherwise we dont pick 2 and we pick >!1. In the case which only one of 1, 2 >!are in A(say x is that number), if when >!adding k to A, the sum becomes a >!square we can take x and put it in the >!set where the other of 1,2 is if that set >!becomes a square we can just add >!the other number back to A. Now if >!both of 1,2 are in the same set other >!than A, if when adding k to A its sum >!becomes a square we can give 1 to A >!if the other set becomes a square too >!we can switch 1 and 2. Now the last >!case: if 1,2 are in separate sets other >!than A, we can do the above >!procedures for another set like B >!because this case ia covered. This >!solution (all of it) depends on T1.

I hope my answer is atleast partially true.

Well i personally believe that:

This procedure can be done for all n. Correct me if im wrong please.

Something we will need later:

T1: Two perfect squares differ by atleast 3. Proof: x >=1: (x+1)^2-x^2= 2x+1 >= 3

Ok the base my proof is:

I will use induction on n. The induction base can be n=9 as in the question. In the induction step we know that the hypothesis is true for n=k-1 now we need to prove the hypothesis for n =k.

Ok now the proof begins:

Now we construct our 3 sets for n=k-1. Now we name the sets :A,B,C. Now if the numbers 2 and 1 are in A and we add k to A, and S(A) becomes a square according to T1 if we take 2 or 1 S(A) will be not a squre and if we add for exmaple 2 to another set and it doesnt become a square then its ok otherwise we dont pick 2 and we pick 1. In the case which only one of 1, 2 are in A(say x is that number), if when adding k to A, the sum becomes a square we can take x and put it in the set where the other of 1,2 is if that set becomes a square we can just add the other number back to A. Now if both of 1,2 are in the same set other than A, if when adding k to A its sum becomes a square we can give 1 to A if the other set becomes a square too we can switch 1 and 2. Now the last case: if 1,2 are in separate sets other than A, we can do the above procedures for another set like B because this case ia covered. This solution (all of it) depends on T1.

I hope my answer is atleast partially true.

Well i personally believe that:

! This procedure can be done for all n. Correct me if im wrong please. Something we will need later:

T1: Two perfect squares differ by atleast 3. Proof: x >=1: (x+1)^2-x^2= 2x+1 >= 3

Ok the base my proof is:

! I will use induction on n. The induction base can be n=9 as in the question. In the induction step we know that the hypothesis is true for n=k-1 now we need to prove the hypothesis for n =k. Ok now the proof begins:

Now we construct our 3 sets for n=k-1. Now we name the sets :A,B,C. Now if the numbers 2 and 1 are in A and we add k to A, and S(A) becomes a square according to T1 if we take 2 or 1 S(A) will be not a squre and if we add for exmaple 2 to another set and it doesnt become a square then its ok otherwise we dont pick 2 and we pick 1. In the case which only one of 1, 2 are in A(say x is that number), if when adding k to A, the sum becomes a squre we can take x and put it in the set where the other of 1,2 is if that set becomes a square we can just add the other number back to A. Now if both of 1,2 are in the same set other than A, if when adding k to A its sum becomes a square we can give 1 to A if the other set becomes a square too we can switch 1 and 2. Now the last case: if 1,2 are in separate sets other than A, we can do the above procedures for another set like B because this case ia covered. This solution (all of it) depends on T1.

I hope my answer is atleast partially true.

Well i personally believe that:

This procedure can be done for all n. >!Correct me if im wrong please.

Something we will need later:

T1: Two perfect squares differ by >!atleast 3. Proof: x >=1: (x+1)^2-x^2= >!2x+1 >= 3

Ok the base my proof is:

I will use induction on n. The >!induction base can be n=9 as in the >!question. In the induction step we >!know that the hypothesis is true for >!n=k-1 now we need to prove the >!hypothesis for n =k.

Ok now the proof begins:

Now we construct our 3 sets for >!n=k-1. Now we name the sets :A,B,C. >!Now if the numbers 2 and 1 are in A >!and we add k to A, and S(A) becomes >!a square according to T1 if we take 2 >!or 1 S(A) will be not a squre and if we >!add for exmaple 2 to another set and >!it doesnt become a square then its ok >!otherwise we dont pick 2 and we pick >!1. In the case which only one of 1, 2 >!are in A(say x is that number), if when >!adding k to A, the sum becomes a >!square we can take x and put it in the >!set where the other of 1,2 is if that set >!becomes a square we can just add >!the other number back to A. Now if >!both of 1,2 are in the same set other >!than A, if when adding k to A its sum >!becomes a square we can give 1 to A >!if the other set becomes a square too >!we can switch 1 and 2. Now the last >!case: if 1,2 are in separate sets other >!than A, we can do the above >!procedures for another set like B >!because this case ia covered. This >!solution (all of it) depends on T1.

I hope my answer is atleast partially true.

Well i personally believe that:

! This procedure can be done for all n. Correct me if im wrong please. Something we will need later:

T1: Two perfect squares differ by atleast 3. Proof: x>=1: (x+1)^2-x^2= 2x+1 >= 3

Ok the base my proof is:

! I will use induction on n. The induction base can be n=9 as in the question. In the induction step we know that the hypothesis is true for n=k-1 now we need to prove the hypothesis for n =k. Ok now the proof begins:

Now we construct our 3 sets for n=k-1. Now we name the sets :A,B,C. Now if the numbers 2 and 1 are in A and we add k to A, and S(A) becomes a square according to T1 if we take 2 or 1 S(A) will be not a squre and if we add for exmaple 2 to another set and it doesnt become a square then its ok otherwise we dont pick 2 and we pick 1. In the case which only one of 1, 2 are in A(say x is that number), if when adding k to A, the sum becomes a squre we can take x and put it in the set where the other of 1,2 is if that set becomes a square we can just add the other number back to A. Now if both of 1,2 are in the same set other than A, if when adding k to A its sum becomes a square we can give 1 to A if the other set becomes a square too we can switch 1 and 2. Now the last case: if 1,2 are in separate sets other than A, we can do the above procedures for another set like B because this case ia covered. This solution (all of it) depends on T1.

I hope my answer is atleast partially true.

Well i personally believe that:

! This procedure can be done for all n. Correct me if im wrong please. Something we will need later:

T1: Two perfect squares differ by atleast 3. Proof: x >=1: (x+1)^2-x^2= 2x+1 >= 3

Ok the base my proof is:

! I will use induction on n. The induction base can be n=9 as in the question. In the induction step we know that the hypothesis is true for n=k-1 now we need to prove the hypothesis for n =k. Ok now the proof begins:

Now we construct our 3 sets for n=k-1. Now we name the sets :A,B,C. Now if the numbers 2 and 1 are in A and we add k to A, and S(A) becomes a square according to T1 if we take 2 or 1 S(A) will be not a squre and if we add for exmaple 2 to another set and it doesnt become a square then its ok otherwise we dont pick 2 and we pick 1. In the case which only one of 1, 2 are in A(say x is that number), if when adding k to A, the sum becomes a squre we can take x and put it in the set where the other of 1,2 is if that set becomes a square we can just add the other number back to A. Now if both of 1,2 are in the same set other than A, if when adding k to A its sum becomes a square we can give 1 to A if the other set becomes a square too we can switch 1 and 2. Now the last case: if 1,2 are in separate sets other than A, we can do the above procedures for another set like B because this case ia covered. This solution (all of it) depends on T1.

I hope my answer is atleast partially true.