Is there a,

$$a^4+(a+d)^4+(a+2d)^4+(a+3d)^4+\dots = z^4\tag1$$

in non-zero integers? One can be familiar with,

$$31^3+33^3+35^3+37^3+39^3+41^3 = 66^3\tag2$$

I found that,

$$29^4+31^4+33^4+35^4+\dots+155^4 = 96104^2\tag3$$

which has $m=64$ addends. The equation,

$$a^4+(a+b)^4+(a+2b)^4+\dots+(a+63b)^4 = y^n\tag4$$


$$64 a^4 + 8064 a^3 b + 512064 a^2b^2 + 16257024 a b^3 + 206447136 b^4 = y^n\tag5$$

for $n=2$ can be reduced to an elliptic curve, so there is an infinite number of primitive integer solutions. However, for $n=4$, it is now a superelliptic curve, so is harder to solve.

Update: Courtesy of Antony's answer, a little experimentation showed that,

$$(\text{Excluding}\; 19^4):\quad 5^4+6^4+7^4+\dots+38^4 = 64^4$$

So close! Perhaps there is a non-zero solution to $(1)$ yet.


  1. What is the general formula for $m$ addends of,$$F(k)=a^k+(a+b)^k+(a+2b)^k+(a+3b)^k+\dots$$ for $k=4?$ (The case $k=2,3$ can be found here, and the special case $a=b=1$ is given by Faulhaber's formula.)

  2. For some $m$ addends, does $F(4)=y^4$ have a solution in non-zero integers $a,b,y$?

  • $\begingroup$ I had already asked a more generalized version of this. If anybody answers this, could you please comment on my question also. math.stackexchange.com/questions/1067581/… $\endgroup$ – ghosts_in_the_code Dec 17 '14 at 19:25
  • $\begingroup$ Okay, I've answered your question. I hope I addressed some of the points you wanted. $\endgroup$ – Tito Piezas III Dec 18 '14 at 0:35

Let, $$S(n)=a^4+(a+d)^4+\dots+(a+nd)^4\tag1$$

Then we can write $$S(n) =a^4+(a^4+4a^3d+6a^2d^2+4ad^3+d^4)+\dots+(a^4+4a^3dn+6a^2d^2n^2+4ad^3n^3+d^4n^4)\\ =(n+1)a^4+4a^3d(1+2+\dots+n)+6a^2d^2(1^2+2^2+\dots+n^2)+4ad^3(1^3+2^3+\dots+n^3)+d^4(1^4+2^4+\dots+n^4)$$

Using Faulhaber's formula for $F(x) = 1^x+2^x+\dots+n^x$ we get $$S(n)=(n+1)a^4+4a^3d \cdot F(1)+6a^2d^2 \cdot F(2)+4ad^3\cdot F(3)+d^4 \cdot F(4)$$

Explicitly $$S(n)=(n+1)\big(a^4+2a^3dn+a^2d^2n(2n+1)+ad^3n^2(n+1)+\tfrac{1}{30}d^4n(2n+1)(3n^2+3n-1)\big)\tag2$$

In general, let $$s(n)=a^x+(a+d)^x+\dots+(a+nd)^x\\ =a^x+(a^x+C^1_xa^{x-1}d+\dots+d^x)+\dots+\big(a^x+C_x^1a^{x-1}dn+\dots+(dn)^x\big)\\ =(n+1)a^x+C_x^1a^{x-1}d(1+2+\dots+n)+C_x^2a^{x-2}d^2(1^2+2^2+\dots+n^2)+\dots+d^x(1^x+2^x+\dots+n^x)\\ =(n+1)a^x+C_x^1a^{x-1}d\cdot F(1)+C_x^2a^{x-2}d^2\cdot F(2)+\dots+d^x\cdot F(x)$$

So $$s(n)=(n+1)a^x+\sum_{i=1}^x C_x^ia^{x-i}d^iF(i)$$

  • $\begingroup$ Thanks! Very nice, using a general approach. Gave $+1$. (I've edited the LaTex format of your answer a bit so it will look clearer.) I've verified your formula $(2)$ gives the correct result when $a,d,n = 29,2,63$. $\endgroup$ – Tito Piezas III Dec 24 '14 at 16:57

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