SOLUTION: hence,prove algebraically that the sum of any two consecutive terms is a perfect square 3;x;10;y;21

You start your question with "hence", so we cannot be sure
what went before that, and how you are supposed to come up
with x and y.  I observe that the three numbers given 3,10,21 
appear in Pascal's triangle 
      
              1
            1   1
          1   2   1
        1   3    3   1
      1   4   6   4   1
    1   5   10  10   5   1
  1   6  15  20  15   6   1
1   7   21  35  35  21   7   1

Pascal's triangle is composed of binomial coefficients which 
are combinations: 

3 = C(3,2), 10 = C(5,2), 21 = C(7,2)

So I assume that the sequence 3,x,10,y,21 is this sequence:

3=C(3,2), x=C(4,2), 10=C(5,2), y=C(6,2), 21=C(7,2) 

,,,,

So the nth term is , and they are called
"triangular" numbers, because they are the numbers of dots 
that can be formed into a triangular arraylike this:                                                         .
                                              .               . .
                              .              . .             . . .
                .            . .            . . .           . . . .
     .         . .          . . .          . . . .         . . . . .
3 = . .   6 = . . .,  10 = . . . .,  15 = . . . . ., 21 = . . . . . .

We need to prove that the sum of any two consecutive terms of this sequence
3,6,10,15,21,... is a perfect square.  We can see that 3+6=9=3², 6+10=16=4²,
10+15=25=5², 15+21=36=6².  We need to prove this in general:

Proof:

the nth term is  and 
the (n+1)st term is  = 

Add them:




which is a perfect square.

Edwin