How can i prove the statement that if the difference of cubes of two consecutive integers is an integral power of 2, then the integer with power 2 can be written as the sum of squares of two different integers.

For example:

$$8^3 - 7^3 = 13^2 = 12^2 + 5^2$$

Any help appreciated.


  • 3
    $\begingroup$ This is true of any odd square... See en.wikipedia.org/wiki/… $\endgroup$ – Brandon Carter Jun 21 '13 at 20:58
  • $\begingroup$ @BrandonCarter i want to know how can i prove this. $\endgroup$ – Shobhit Jun 21 '13 at 21:01
  • $\begingroup$ @BrandonCarter. Maybe I mis-understand you. None of $9,49, 81,...$ etc., is the sum of two squares.If $(n+1)^3=n^3=m^2>1$ then what we need to show that $m$ has a prime factor congruent to $1 \mod 4.$ $\endgroup$ – DanielWainfleet Jul 17 '18 at 2:58

We can see that $$n^3-(n-1)^3=3n^2-3n+1.$$ By looking at congruences modulo 4, we see that $$n^3-(n-1)^3 \equiv 1\pmod 4.$$ Recalling that you assumed that the difference was a square, you can apply Fermat's theorem cited above to conclude.


Im pretty sure youve heard of an identity $$ a^3-b^3=(a-b)*(a^2 + b^2 + ab) $$
If the above is a perfect square then it must be of form $a+b=n$ and $a^2 + b^2 + ab = n x^2$.

Now you can proceed. I think this method is good for an approach.


By expanding, we get $x^2=3n^2+3n+1$ Thus $4x^2-1=12n^2+12n+3=3(2n+1)^2=(2x-1)(2x+1)$ $gcd(2x-1,2x+1)=1$, We have two cases. Case 1: $2x-1$ can be written as $3k^2$ Hence 2x+1 is a perfect square and 3|(2x+1)-2. No solution. Therefore $2x-1$ must be a perfect square. Let it be $(2a+1)^2$. Then $x=2a^2+2a+1=a^2+(a+1)^2$ QED

Note: I came up with this solution myself, but it also turned out to be the official solution for the IMO Longlist problem.


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