Question

4 numbers form an increasing AP one of these numbers is sum of squares of other three find the numbers

Answer 1

Hello.
Let's begin solving.

Suppose we have the AP: 
a-d, a, a+d, a+2d 

Then for the last term to = ∑(squares of other 3), we would have: 

3a^2 + 2d^2 = a + 2d 
3a^2 - a + 2d(d - 1) = 0 

Solve for a: 

a = [1 ±√(1 - 24d(d-1))]/6 

Now for the sequence to be all integers, a and d must be integers. 
For it to be increasing, d ≥ 1 
But that means that to get a real number for a, the discriminant up there, 
∆ = 1 - 24d(d-1), 
must be ≥0, which can happen only if d = 1. Because d ≥ 1, and any integer > 1 will make ∆ < 0, making "a" complex. 
So with d = 1, a = (1 ± 1)/6, and of these, only a = 0 is an integer. So: 
a = 0; d = 1, 
and the sequence has to be: 
a-d, a, a+d, a+2d = -1, 0, 1, 2 
... for the last term to be the one that's ∑squares. 

But no other term can take this role, because it will either have to be negative, and therefore unable to be a sum of squares, or else the term following it will have greater magnitude than it has, and that's impossible -- for an integer to be greater than its own square, let alone its own plus some others. 

So that's the only solution.
Hope it helps