3
$\begingroup$

I read something about how the possible digits of any square number is $0,1,4,5,6,9$ and that I could bash it out, but then that would take a lot of time. Is there any other way to to this problem?

$\endgroup$
  • 1
    $\begingroup$ What do you know about Pythagorean Triples? $\endgroup$ – Brevan Ellefsen Mar 25 '17 at 3:09
  • $\begingroup$ The possible final digits can only be those digits. $\endgroup$ – Thomas Andrews Mar 25 '17 at 3:09
  • $\begingroup$ Pythagorean Triples are integers $a,b,c$ in the form $a^2$+$b^2$=$c^2$ $\endgroup$ – jonyoung2002 Mar 25 '17 at 3:13
  • 1
    $\begingroup$ OK. Looked it up. So what I would do is first write 2017 as the sum as 2 squares (9,44) and substitute those values for m and n. Then I would do $44^2$-$9^2$=a and $44$*$9$*2=b. $\endgroup$ – jonyoung2002 Mar 25 '17 at 3:20
  • 1
    $\begingroup$ Got it! Thanks! $\endgroup$ – jonyoung2002 Mar 25 '17 at 3:21
4
$\begingroup$

By combining Fermat's theorem on sums of squares and the irreducible elements in the Gaussian integers $\mathbb{Z}[i]$ one can find the number of representations of an integer n(not necessarily a square) when in n's prime factorization all primes are congruent to 1 mod 4. We have the following method which gives the number of the representations and how to find them. Let:

$$n=p_{1}^{a_{1}}...p_{n}^{a_{n}}$$

Such that all $p_{k}$ are congruent to 1 mod 4. There are exactly $4(a_{1}+1)...(a_{n}+1)$ representations of n as a sum of squares. And these can be found by factorizing $p_{k}$ in the Gaussian integers as $p_{k}=(a+bi)(a-bi)$ and multiplying any combination made by taking $a_{n}$ guys from each $p_{k}$'s factorization.

For your specific problem, since 2017 is a prime congruent to 1 mod 4 by the formula above it can be written in 4x3=12 ways. 2017's factorization is: $$2017=(44+9i)(44-9i)$$ hence $$2017^2=(44+9i)^2(44-9i)^2$$ by picking any two guys(we use two picks since the others are the same) from the above representation you get all the answers except for changing signs and summands: $$1855+792i=(44+9i)(44+9i)$$ $$2017+(0)i=(44+9i)(44-9i)$$ So 6 of 12 solutions with all possible choices of signs are $2017^2=(\pm1855)^2+(\mp792)^2=(\pm2017)^2+0^2$. The other 6 solutions are made by changing the order of summands.

Note factorizing 2017 in the Gaussian integers is the same as finding the only two numbers such that $2017=a^2+b^2=44^{2}+9^{2}$ which still takes time since 2017 is a large prime, but is still easier than doing the same directly for 2017^2. Generally this method works better for small primes in factorizations.

$\endgroup$
2
$\begingroup$

All Pythagorean triples $(a,b,c)$, $a^{2}+b^{2}=c^{2}$, satisfy the formulas \begin{equation*} a=k(m^{2}-n^{2}),\quad b=2kmn,\quad c=k(m^{2}+n^{2}),\qquad m>n> 0,\quad k> 0. \end{equation*}

Since $c=2017=44^{2}+9^{2}$(see this answer), $k=1$. Then $m=44,n=9$, and \begin{eqnarray*} a &=&m^{2}-n^{2}=44^{2}-9^{2}= 1855, \\ b &=&2mn= 2(44)(9)= 792. \end{eqnarray*} The other solutions are found by allowing negative values or swapping $ a $ with $ b $. So $ a=\pm 1855, b=\pm 792$, $ a=\pm 792, b=\pm1855 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.