There are 4 terms of an increasing arithmetic progression (AP) where one term is equal to the sum of the squares of the other three.

Question

If there are 4 terms of an increasing arithmetic progression (AP) where one term is equal to the sum of the squares of the other three, find the terms.

I have tried solving the question by taking the terms as $a-3d$, $a-d$, $a+d$ & $a+3d$. I am getting $a=-0.5$ but $d$ is coming out to be a complex number, which shouldn't happen.

Answer 1

Start with any arithmetic progression $a,a+d,a+2d,a+3d$. Select any term as the one that will equal the sum of the squares of the other three: say, $a+d$. Now solve for $k$: $$k(a+d)=k^2(a^2 + (a+2d)^2+(a+3d)^2)$$ In other words, take $$k=\frac{a+d}{a^2 + (a+2d)^2+(a+3d)^2}$$ Now, by construction, the arithmetic progression $$ka,k(a+d),k(a+2d),k(a+3d)$$ satisfies your conditions.

It should be clear to you, therefore, that there are many solutions to your problem. Perhaps you forgot to tell us that the terms must be integers?

Answer 2

• Case where the biggest term is the sum of the squares of the three others.

The 4 terms are $a-r,a,a+r$ and $a+2r$.

We need $a+2r=(a-r)^2+a^2+(a+r)^2=3a^2+2r^2$,

thus $3a^2-a+2r(r-1)=0$, u.c. $r>0$

This quadratic equation has a pair of real solutions iff $\Delta=1-24r(r-1)>0$, hence $r(r-1)<1/24$

Solving in turn $r^2-r-1/24=0$, we get $\Delta_r=7/6>0$, so there are two solutions $r_1<r_2$ and the acceptable region for $r$ is $[r_1,r_2]$.

For instance, for $r=1$, $\Delta=1$, $a_1=0, a_2=1/3$ and you get the solutions:

$\{-1,0,1,2\}$ : indeed, $(-1)^2+0^2+1^2=2$

$\{-2/3,1/3,4/3,7/3\}$ : indeed, $(-2/3)^2+(1/3)^2+(4/3)^2=\frac{4+1+16}{9}=21/9=7/3$

When $r$ varies you have an infinity of solution-quadruplets, and then you can also check the three other cases that will probably bring in more solutions.

Edit: If you have the further constraint that terms must be integers, then since $0$ and $1$ are the only integers in $[r_1,r_2]$, you can only pick $r=1$ and the two quadruplets above to solve the first case.

Second edit:

• Case where the smallest term is the sum of the squares of the three others.

The 4 terms are $a-r,a,a+r$ and $a-2r$.

We need $a-2r=(a-r)^2+a^2+(a+r)^2=3a^2+2r^2$,

thus $3a^2-a+2r(r+1)=0$, u.c. $r>0$

This quadratic equation has a pair of real solutions iff $\Delta=1-24r(r+1)>0$, hence $r(r+1)<1/24$

And you can pick any $r$ between $0$ and $\frac{\sqrt{7/6}-1}{2}$.

Answer 3

Using your values for the AP and guessing that the last term is the sum of the squares of the rest, I get $$a+3d=(a-3d)^2+(a-d)^2+(a+d)^2\\ =3a^2-6ad+11d^2\\ a=\frac 16(6d+1\pm \sqrt{(6d+1)^2-44d^2})\\ =\frac 16(6d+1\pm\sqrt{1+24d-8d^2})$$ which is nicely real as long as $d$ is not too big. I don't find a unique solution.