It is well known that every positive integer is the sum of at most four perfect squares (including $1$).

But which positive integers are not the sum of four non-zero perfect squares ($1$ is still allowed as a perfect square) ?

I showed that the numbers $2^k$ , $2^k\cdot 3$ and $2^k\cdot 7$ with odd positive integer $k$ have this property. I checked the numbers upto $10^4$ and above $41$, no examples , other than those of the mentioned forms , occured. So my question is whether additional positive integers with the desired property exist.

  • $\begingroup$ Another formulation of the question : If $n>41$ is an integer and $n=a^2+b^2+c^2+d^2$ has no solution in positive integers, must $n$ be of the form $2^k$ or $2^k\cdot 3$ or $2^k\cdot 7$ with odd positive integer $k$ ? $\endgroup$
    – Peter
    Commented Jun 9, 2018 at 13:51
  • $\begingroup$ Worth mentioning is OEIS sequence A000534 "Numbers that are not the sum of 4 nonzero squares." $\endgroup$
    – Somos
    Commented Apr 2, 2020 at 21:03

2 Answers 2


page 140 in Conway's little book, $$ 1,3,5,9,11,17,29,41, \; 2 \cdot 4^m \; , \; 6 \cdot 4^m \; , \; 14 \cdot 4^m \; . $$ The proof is on the same page, with preparatory material in the previous few pages.

The first detail: any number $3 \pmod 8$ is the sum of three squares, meanwhile they must be odd squares, therefore nonzero. The square of any number that is divisible by $4$ becomes $0 \pmod 8.$ As a result, any number $6 \pmod 8$ is the sum of three squares, as $ (2A)^2 + B^2 + C^2,$ where $A,B,C$ must be odd squares, therefore nonzero.

10 June: Second detail: if $x^2 + y^2 + z^2 \equiv 0 \pmod 4,$ then $x,y,z$ are all even. This means that $12 \pmod{32}$ is the sum of three nonzero squares. Same for $24 \pmod{32}$

  • $\begingroup$ @JamesArathoon I transcribed incorrectly. I put in a link for a pdf, you can check page 140 yourself. Believe I have it now. $\endgroup$
    – Will Jagy
    Commented Jun 9, 2018 at 18:17
  • $\begingroup$ Thanks for correcting transcription. All the primes listed are the sum of 2 or 3 non-zero squares. In the case of the largest 3 primes above, 17, 29, and 41 they are the sum of both 2 and 3 non-zero squares. The proof shows that any prime larger than 41 must be the sum of 4 non-zero squares, even if that prime is the sum of both 2 and 3 non-zero squares. $\endgroup$ Commented Jun 9, 2018 at 18:34
  • $\begingroup$ Thank you. So every other number can be expressed as the sum of four non-zero squares. $\endgroup$
    – Peter
    Commented Jun 9, 2018 at 20:07
  • $\begingroup$ @Peter yes. I put a link to a pdf of Conway's book. You would like the first chapter, he defines the Topograph method, which becomes especially useful when trying to find all solutions $(x,y)$ to $A x^2 + B xy + C y^2 = N.$ Recently you were asking about finding all solutions to $x^2 - k y^2 = n,$ the topograph is the responsible way to do that. $\endgroup$
    – Will Jagy
    Commented Jun 9, 2018 at 20:14
  • $\begingroup$ @WillJagy Sorry, I looked at page $140$ , but I didn't find where the key for the proof is. Please help! $\endgroup$
    – Peter
    Commented Jun 9, 2018 at 20:38

Some of my topograph answers, in order by question number. I got better with the diagrams as time went by. If you just look at these, not much will happen. If you draw some of your own examples, you will begin to understand.



http://www.maths.ed.ac.uk/~aar/papers/conwaysens.pdf (Conway)

http://www.springer.com/us/book/9780387955872 (Stillwell)

https://www.math.cornell.edu/~hatcher/TN/TNbook.pdf (Hatcher)

http://bookstore.ams.org/mbk-105/ (Weissman)









http://math.stackexchange.com/questions/739752/how-to-solve-binary-form-ax2bxycy2-m-for-integer-and-rational-x-y/739765#739765 :::: 69 55





http://math.stackexchange.com/questions/1132187/solve-the-following-equation-for-x-and-y/1132347#1132347 <1,-1,-1>







http://math.stackexchange.com/questions/1737385/if-d1-is-a-squarefree-integer-show-that-x2-dy2-c-gives-some-bounds-i/1737824#1737824 "seeds"


Is there a simple proof that if $(b-a)(b+a) = ab - 1$, then $a, b$ must be Fibonacci numbers? 1,1,-1; 1,11

To find all integral solutions of $3x^2 - 4y^2 = 11$

  • $\begingroup$ Excellent! will help me a lot! $\endgroup$
    – Michael
    Commented Jun 11, 2018 at 22:58

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