Put 4 different numbers in the circles in the corner of a sq

Put 4 different numbers in the circles in the corner of a square so that when you add the numbers at the end of each line you always get a square #.

Solution to the puzzle:

Let A, B, C, D be the 4 corners of the square.

(a)

1 , 3 , 46 and 35 respectively be the numbers placed at A, B, C and D :

On the line AB, 1+3=2^2.

On BC, 3+46=7^2 .

On CA, 46+35=9^2 and

on AD, 1+35=6^2.

So,1 , 3 , 46 and 35 is a solution as they are diffrent numbers and on each line we saw that the sum is equal to a square.

Now let us consider 5, 4 , 117 and 139 respectively at A,B,C and D.

5+4=9=3^2 ,

4+117=121= 11^2 ,

117+139=256= 16^2 and

139+5 = 12^2 .

Therefore, the numbers 5,4,117 and 139 at A,B,C and D makes a solution. They are different and their sum is a square on each line .

1 , 3, 118 , 323 be the numbers respectively at A,B,C and D .

1+3=4= 2^2 ,

3+118 = 11^1 ,

118+323=441= 21^2 and

323+1=324= 18^2 .

Therefore, 1 at A , 3 at B, 118 at C and 323 at D satisfies the condition of getting perfect squares by totalling on each sides .

Any cyclic permutation of a solution is also a solution.