If a,b,c,d are distinct integers form an increasing AP such that d=a^2 +b^2 +c^2,then find a+b+c+d.

Question

This is the qustion in AP which i am unable to solve after trying many times.after using d=a²+b²+c²  and ap structure i am getting a form of quadratic equation .beyond that i fail.

Answer 1

Hi again Arpan,

 

Let the common difference be x.

 

(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)

                    =d+d^2+2((d-3x)(d-2x)+(d-3x)(d-x)+......+(d-x)d)

 

Now first of all, we have used all the information and secondly we are expecting a solution.  Given the

RHS is quadratic in both d and x, we must expect it to be a perfect square of the form (d-kx)^2, k an integer and the solution will be one of the terms in the AP.  One or more of a,b,c must be negative.  This is my intuition.  However, the d term will not allow this to be a perfect square so probably this is a fruitless approach yielding a Diophantine quadratic which probably only yields to guessing.

 

We need another approach.  The idea that one or more of a,b,c is negative is still valid...let's see how:

x>0 because d=a^2+b^2+c^2 > c 

 

a^2+b^2+c^2=3a^2+5x^2+6ax=d=a+3x

So 3a^2+5x^2=a+3x-6ax    (you can probably see the answer by the coefficients 3+5=?1?3?6  but let's suspend that thought)

 

3a^2+5x^2=3x(1-2a)+a  now a is not 0 because this leads to nonsense equation in x.

Now a is an integer by definition...but so must x be an integer >=1, otherwise the next integer in the series would not be.

 

The LHS is positive.  The RHS is a-3kx where k>=a for all a,x>=1. So the RHS is negative for all a>=1

Therefore a<0

 

3a^2+5x^2=a+3x-6ax

is a Diophantine quadratic.  

 

I find it intractable algebraically , but 3+5=-1+3+6 looking at the coefficients.

So by inspection a=-1 x=1

 

(a,b,c,d)=(-1,0,1,2)  so a+b+c+d=2

 

Check: a^2+b^2+c^c=1+0+1=2=d OK