Question 1104164
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For  "c"  be divisible by  3,  the necessary and sufficient condition is  {{{c^2}}}  is divisible by  3.


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The Table below contains two inputs:

    a) vertical column for "a mod 3"  (leftmost column),  and

    b) horizontal row  for "b mod 3"  (uppermost row).


Next vertical column is "a^2 mod 3".

Next horizontal row  is "b^2 mod 3".


The table itself contains the values  "{{{a^2 + b^2}}} mod 3"  in its cells.


                             0    1    2         <<<---===  b   mod 3

   a mod 3   {{{a^2}}} mod 3        0    1    1         <<<---===  {{{b^2}}} mod 3

     0          0            0    1    1

     1          1            1    2    2

     2          1            1    2    2


From the table, you can see that the sum  {{{a^2+b^2}}}  is multiple of 3 if and only if both "a" and "b" are multiples of 3.


Then and only then you have  "0 mod 3" in the table.



Now the proof is this chain of arguments:


    If "c" is multiple of  3,  then  {{{c^2}}}  is multiple of 3,   and  since {{{c^2}}} = {{{a^2 + b^2}}},  

    it is possible if and only if  both "a"  and  "b"  are multiplies of 3.
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It is the shortest way to prove the statement.