4
$\begingroup$

I'm studying about number theory by myself. So I'm sorry if the question seems to be unclear.

I just want to know how to prove that one member of a Pythagorean triple is always divisible by 5 and that the area of an integer-sided right-angled triangle is always divisible by 6.

I searched for similar questions but didn't find an understandable proof for me.

$\endgroup$
  • $\begingroup$ cut-the-knot.org/pythagoras/pythTripleDiv.shtml $\endgroup$ – lab bhattacharjee Oct 7 '16 at 17:30
  • $\begingroup$ I have already seen this but I'm still not getting it. $\endgroup$ – Ahmed Amir Oct 7 '16 at 17:45
  • $\begingroup$ Tell us what confuses you about the linked proof so we can help you work through it. $\endgroup$ – rogerl Oct 7 '16 at 17:52
  • $\begingroup$ Literally, it's not clear for me. I need an explanation for this in words if it's possible. $\endgroup$ – Ahmed Amir Oct 7 '16 at 18:09
4
$\begingroup$

We are considering integer solutions to $$ a^2+b^2=c^2 $$


Let us consider the equation $a^2+b^2=c^2$ modulo $2$: $$ \begin{array}{c|cc} +&0^2&1^2\\ \hline 0^2&0^2&1^2\\ 1^2&1^2&0^2 \end{array} $$ we see that in all four cases at least one of the three numbers is zero modulo $2$. So at least one in the triple is divisible by $2$.


Modulo $3$ we have $1^2=2^2$ so the same table applies. At least one is divisible by $3$.


Modulo $4$ we have $0^2=2^2$ and $1^2=3^2$ so again the same table applies. At least one is divisible by $4$.


Modulo $5$ we have $1^2=4^2=1$ and $2^2=3^2=4$ and so $$ \begin{array}{c|ccc} +&0^2&1^2&2^2\\ \hline 0^2&0^2&1^2&2^2\\ 1^2&1^2&\times&0^2\\ 2^2&2^2&0^2&\times \end{array} $$ where $\times$ indicates that result cannot be a square, since $1^2+1^2=2$ and $2^2+2^2=3$ are not values of squares modulo $5$. They are quadratic non-residues mod $5$.

All $7$ cases that are actually possible contain at least one zero. So at least one of the three numbers is divisible by $5$.


Finally, regarding the area being divisible by $6$, we see from the formulas for generating the triples $$ (a,b,c)=(m^2-n^2,2mn,m^2+n^2) $$ that the area must be $$ T=\tfrac12 ab=(m^2-n^2)mn $$ and if neither $m$ nor $n$ is divisible by $2$ we have $m=n=1$ modulo $2$ and therefore $m^2-n^2$ must be divisible by $2$.

If neither $m$ nor $n$ is divisible by $3$, we see that either $m,n\in\{1,2\}$ modulo $3$ so that $m^2-n^2=0$ modulo $3$. So indeed the area is divisible by both $2$ and $3$, and hence by $6$.

$\endgroup$
1
$\begingroup$

From the link:

"The case to consider is m = ±1 (mod 5) or m = ±2 (mod 5). And the same is true for n."

Here, we want to ultimately look at the squares, because we have already dealt with m or n being divisible by 5 on the previous condition.

"It then folows that both $m^2$ and $n^2$ (edited here) may only be 1 or 4 modulo 5."

Just do the multiplication and reduce modulo five for all of the cases.

"If they are equal modulo 5, then m2 - n2 = 0 (mod 5). Otherwise, m2 + n2 = 0 (mod 5)."

When the first statement is established and we have reduced to only the case where the squares of the elements can only be 1 or 4 mod 5, then either they are equal or the aren't. If they are, then the subtraction case applies; if not, then the addition one does.

If there is still some confusion, it might be worth looking up modular arithmetic.

$\endgroup$
  • $\begingroup$ Yeah, I'm not familiar with "modular arithmetic". But I have seen an understandable proof shows that a or b should be divisible by 3. math.stackexchange.com/a/1083833/376166 $\endgroup$ – Ahmed Amir Oct 8 '16 at 8:02
  • $\begingroup$ So I want to write a proof for my question using the same strategy in that linked answer (Using K). $\endgroup$ – Ahmed Amir Oct 8 '16 at 8:05
0
$\begingroup$

As I said, I haven't learned about "modular arithmetic" yet. That's may be why I'm not getting your answers which seem to be obviously clear.

Despite this, I tried to prove it on my "Elementary" way and that was what I came up with.

"one member of a Pythagorean triple is always divisible by 5"

First, every integer (m, n) can be written on one of these forms (5k, 5k+1, 5k+2, 5k+3, 5k+4). We take the last four forms and square them. We will end only with two different forms(5k+1, 5k+4). Note that: (5k+9) can be rewritten to be (5k+4) with a different value of k.

If m^2 and n^2 can be written on the same form (5k+1, 5k+1) or (5k+4, 5k+4) then m^2 - n^2 should be divisible by five. If m^2 and n^2 are written on different forms (5k+1, 5k+4) then m^2 +n^2 should be divisible by 5 cause (4 + 1 = 5).

Now we have done for the last four forms and still have the first form which is (5k). Simply, if m can be written on the form (5k) and n is written on any other form then m * n should be divisible by five, Therefore, 2mn also should be divisible by 5.

Note: 1) My English isn't perfect so I'm sorry if there are some mistakes. 2) The proof is clear for me now and I can understand the main parts of your proofs. However, I won't close the question right now. So if you have any comments on what I have written, kindly, inform me.

$\endgroup$

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.