Question

can you find four numbers such that the sum of every two numbers and the sum of all four numbers may be perfect squares​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Take proportionals as a:ax::b:bx

Now three equations are given

 a+bx=21 .....(i)

ax+b=19 .....(ii)

 a2+(ax)2+b2+(bx)2=442 

⇒(a2+b2)(1+x2)=442 .....(iii)

Add (i) and (ii), we get

(a+b)(1+x)=40 ......(iv)

Subtract (ii) from (i), we get

(a−b)(1−x)=2 ......(v)

Multiply (iv) and (v)

(a2−b2)(1−x2)=80 ......(vi)

Now add (iii) and (vi) and get equation 

a2+(bx)2=261 

Now both a2 and (bx)2 are perfect squares. 

Now, we have to split 261 into two parts such that both are perfect squares and it can be done as 36+225.

So, a=6 and bx=15

Now subtract (vi) from (iii), we get

(ax)2+b2=181 

Now ax and b are also perfect square so split 181 in two parts which is 81+100.

So, ax=9 and b=10

So, the proportionals are 6,9,10,15.