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I have a question about factorization of number $3^{12}-6^6+2^{12}$. By completing the square one can show that$$3^{12}-6^6+2^{12} = (3^6-2^6)^2+6^6 = 665^2+216^2$$ If we can find another representation of this number as sum of squares then may try Euler's factorization method. But how to find another sum of squares in the simplest way? Maybe other methods work?

serg_1
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  • There is this method using the triangular numbers to find the sum of two squares through solving a quadratic equation, that is there is no need to factorize the original number. https://math.stackexchange.com/questions/1972771/is-this-the-general-solution-of-finding-the-two-original-squares-that-add-up-to – user25406 Oct 08 '21 at 12:38
  • @user25406 The use of triangle numbers in that post does nothing but to complicate the discussion. They are not used in an essential way, Additionally, there was an assertion in the text that seemed to assert that this method can always be used to express a number as a sum of two squares, which is false, causing me to doubt the validity of the rest. Simplifying notation, you are solving a quadratic to find when, for a fixed k, you can write $N=n^2+(n+k)^2$, and the answer is "not usually". The condition the the discriminant is a square is equivalent to expressing $2N$ as a sum of two square. – Aaron Oct 08 '21 at 16:05
  • @user25406 (continued) The solution proposed to find a working $k$ is to simply go through all the values of $k$ that work, which is simply brute forcing the problem, not using anything that simplifies things. It is not an adequate method. – Aaron Oct 08 '21 at 16:07
  • serg_1, where did you get the problem? What methods are they discussing in that chapter? – Will Jagy Oct 08 '21 at 16:08
  • By the way, there are two more decompositions: $488881=145^2+684^2=359^2+600^2$. – user25406 Oct 08 '21 at 19:38
  • @Aaron, how is $488881= 216^2 + (216+449)^2$ wrong? – user25406 Oct 08 '21 at 19:46
  • @user25406 I never said that a specific decomposition was incorrect (not that you had even given one when I made my criticism), I said that your method is not an effective algorithm to find them. at least as described. It reduces one problem to an equally hard one, namely that of finding a value of $k$. If you are claiming that you found the decompositions with your method and that you didn't have to check tons of different values of k, I would appreciate seeing that computation (and not just "take these 4 values of k") – Aaron Oct 08 '21 at 20:11
  • @Will Jagy, it's an ordinary hometask of factorization from the number theory book of: $3^{18}+2^{18}$ which is $(2^6+3^6)(3^{12}-6^6+2^{12})$ – serg_1 Oct 09 '21 at 06:36
  • serg, what book would that be? If not in English, perhaps there is a translation. Note that Euler allows, for example, finding two ways of writing your number as $x^2 + 3 y^2.$ It is natural to consider this as your number was given as $u^2 - uv + v^2$ – Will Jagy Oct 09 '21 at 15:38
  • It turns out any prime $p$ dividing your number must be of the form $36n+1,$ that is $p \equiv 1 \pmod 4$ and $p \equiv 1 \pmod 9 ; . ; ; $ The first few such primes are $$37 ; 73 ; 109 ; 181 ; 397 ; 433 ; 541 ; 577 ; 613 ; 757 ; 829 ; 937 ; 1009 ; 1117 ; $$ and trial division by these primes, in order, quickly shows the number to be $37 \cdot 73 \cdot 181 $ – Will Jagy Oct 10 '21 at 16:14
  • Proof of the 1 mod 36 business by Jyrki at https://math.stackexchange.com/questions/4272976/factoring-a-cyclotomic-polynomial-modulo-a-prime – Will Jagy Oct 10 '21 at 19:16

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We have \begin{eqnarray*} 665^2+216^2 = 488881 = 37 \times 13213 = (1^2+6^2)(a^2+b^2)=(a+6b)^2+(6a-b)^2. \end{eqnarray*} Now solve $a+6b=665,6a-b=216$ gives $a=53,b=102$ now swap $a$ and $b$ to get another representation as the sum of two squares.

$$ 420^2+559^2 =488881=665^2+216^2.$$

Donald Splutterwit
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    Since the end goal of the problem seems to be using the sum of squares representations to factor the number, using a factorization seems a bit against the spirit of the question. – Aaron Oct 07 '21 at 21:43
  • @Aaron It seems more difficult to find the representation $420^2+559^2$ and then use this to conclude $X=37 \times 13213$. Than to factorise & figure the other way around? ... Should I delete this answer then ? – Donald Splutterwit Oct 07 '21 at 21:49
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    I would leave the answer, because it still demonstrates something that is interesting and relevant. And I agree that it seems difficult to find other representations without factoring (and, in fact, because of the factorization into primes, one can calculate that there will be 4 ways to write the number as a sum of squares, presumably by using your method and expressing each of the prime factors as a sum of squares). Only OP can really assess if this is giving what they are looking for. – Aaron Oct 07 '21 at 21:55
  • @Aaron, there is a method that finds all the representations of the sum of two squares using the triangular numbers and solving a quadratic equation. The link is given in the above comment. – user25406 Oct 08 '21 at 12:41
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    It turns out that prime factors $p$ must satisfy $p \equiv 1 \pmod{36} ; . ;$ The first few of these are $37, 73, 109, 181, 397,; ; $ and trial division quickly gives $37 \cdot 73 \cdot 181$ – Will Jagy Oct 10 '21 at 20:11