# If distinct integers $a,b,c,d$ are consecutive terms of an arithmetic progression such that . . .

If distinct integers $$a,b,c,d$$ are consecutive terms of an arithmetic progression such that $$a^2 + b^2 + c^2=d$$ then $$a+b+c+d$$ equals? $$(A)\ 0 \quad (B)\ 1 \quad (C)\ 2 \quad (D)\ 4$$ Here is what I tried: assume the common difference is $$x$$. Thus, $$a^2 + a^2 + 2ax + x^2 +a^2 +4ax +4x^2=a + 3x$$ That is, $$3a^2 + 6ax + 2x^2=a + 3x$$ First I assumed it as a quadratic in $$a$$ and solved for it. $$a=\frac{(1-6x) \pm \sqrt{12x^2-48x +1}}{6}$$ but to no avail. Then I assumed it as a quadratic in $$x$$ and got: $$x=\frac{(3-6a) \pm \sqrt{12a^2 -36a +81}}{6}$$ And I noticed that I'm stuck. Please help. Thanks in advance!

Edit: I made a calculation error as pointed out by multiple answers, and so the equation should have been $$3a^2 + 6ax + 5x^2=a + 3x$$ which leads to different solutions.

• If I'm not mistaken, you can't assume that they are one after another in the arithmetic progression, right? You might need to make the terms a bit different. I'm not sure if it's helpful, but it might be better to write the sum as $a^2 + a + b^2 + b + c^2 +c$. Jun 30, 2019 at 10:08
• @CameronWilliams with reference to other problems in the same book, it should mean that they are consecutive terms. Jun 30, 2019 at 10:10
• In that case, I edited your post to make this clear for others! I hope you don't mind! Jun 30, 2019 at 10:21

Hint:

Let $$a,b,c,d$$ be $$A-3D, A-D, A+D, A+3D$$

$$A+3D=(A-3D)^2+(A-D)^2+(A+D)^2=3A^2+11D^2-6AD$$

$$3A^2-A(6D+1)+11D^2-3D=0$$

The discriminant $$(6D+1)^2-12(11D^2-3D)=-96D^2+48D+1=1+6-96\left(D-\dfrac14\right)^2\le7$$ has to be perfect square

• I'm sorry, I could not proceed. Could you please explain the cryptic last line "has too perfect square"? Jun 30, 2019 at 10:21
• @AryanSonwatikar, Has to be perfect square. Jun 30, 2019 at 10:22
• Oh, okay. So I simply need to check for discriminant equals $0,1,4$ right? Jun 30, 2019 at 10:24
• Okay, $D=0.5=A$ which leads me to the answer as option $C$ which matches with the answer key. Thank you so much! Jun 30, 2019 at 10:32
• @AryanSonwatikar, Welcome! Had it been odd number of terms, we would have chosen $$\cdots,A-D,A, A+D,\cdots$$ Jun 30, 2019 at 10:38

Your correct equation is $$3a^2+a(6x-1)+5x^2-3x=0$$ whose discriminant is $$(6x-1)^2-12(5x^2-3x)=1+6-24\left(x-\dfrac12\right)^2\le7$$

or the equation can be formed as $$5x^2+x(6a-3)+3a^2-a=0$$

Calculate the discriminant which needs to be perfect square

You made a slight error, this needs to be $$5x^2$$ and not $$2x^2$$.

So you have this equation $$3a^2+a(6x-1)+(5x^2-3x)=0$$

This equation has discriminant $$\Delta=-24x^2+24x+1$$

The way to solve this for Diophantine equations is to introduce an integer $$m$$ such that $$\Delta=m^2$$ (because we need the square root to be an integer for the solution $$a$$ to have a chance to be an integer).

Now we are left to solve $$24x^2-24x+(m^2-1)=0$$

Once again since $$x$$ is an integer, the discriminant $$\delta=672-96m^2=96(7-m^2)$$ should also be a perfect square.

In this case notice that $$m$$ can only take values $$\ 0,1,2\$$ since we need $$\delta\ge 0$$

$$96=2^5\times 3$$ so we need to multiply by $$6$$ to have perfect square and the solution is $$m=1$$.

Finally $$x=\dfrac{24\pm\sqrt{96\times 6}}{2\times 24}\in\{0,1\}$$

The last step is to verify the $$x$$ found satisfy the original problem.

• $$x=0$$

Then $$a=b=c=d\implies d=3a^2=a\implies a=\frac 13$$ which is not an integer. So this is not a valid solution.

• $$x=1$$

With this value we arrive to $$3a^2+5a+2=(a+1)(3a+2)=0$$

And only $$a=-1$$ is a valid integer.

Conclusion: $$a=-1,\ b=0,\ c=1,\ d=2$$

We can verify $$a^2+b^2+c^2=1+0+1=2=d$$ and $$a+b+c+d=-1+0+1+2=2$$.

Let $$x$$ be the common difference.

From $$a^2+b^2+c^2=d$$, we get $$d > 0$$,

Since $$a,b,c,d$$ are distinct, we get $$x\ne 0$$.

If $$x < 0$$, then $$c > d$$, hence $$c^2 > d^2\ge d=a^2+b^2+c^2 > c^2$$ contradiction.

Hence $$x > 0$$.

\begin{align*} \text{Then}\;\;&a^2+b^2+c^2=d\\[4pt] \implies\;&(b-x)^2+b^2+(b+x)^2=b+2x\\[4pt] \implies\;&3b^2+2x^2=b+2x\\[4pt] \implies\;&b(1-3b)=2x(x-1)\\[4pt] \implies\;&b(1-3b)\ge 0\\[4pt] \implies\;&b=0\;\;\\[4pt] \implies\;&2x(x-1)=0\\[4pt] \implies\;&x=1\\[4pt] \implies\;&a,b,c,d=-1,0,1,2\\[4pt] \implies\;&a+b+c+d=2\\[4pt] \end{align*}