Expert: Paul Klarreich - 3/19/2009

Question
QUESTION: Hi I have a question concerning the Pythagorean triple, which mean that there are three natural numbers x, y, z provided that x<y<z.  I have to assume a, b, and c form a Pythagorean triple and prove that either a is congruent to 0 (mod 5), b is congruent to 0 (mod 5), or c is congruent to 0 (mod 5).

If I work this by cases, there will be 64 cases!  I have already worked a couple (ranging from a=5m+1, b=5n+2, and c=5q+3).  I notice that for each I get

5r=m^2 + n^2 - q^2

How do I show that either a, b, or c is congruent to 0 (mod 5) using this?  Or can I?

ANSWER: Questioner:   Jen
Category:  Advanced Math
Private:  No
 
Subject:  Pythagorean Triple Proof
Question:  Hi I have a question concerning the Pythagorean triple, which mean that there are three natural numbers x, y, z provided that x<y<z.  I have to assume a, b, and c form a Pythagorean triple and prove that either a is congruent to 0 (mod 5), b is congruent to 0 (mod 5), or c is congruent to 0 (mod 5).

If I work this by cases, there will be 64 cases!  I have already worked a couple (ranging from a=5m+1, b=5n+2, and c=5q+3).  I notice that for each I get

5r=m^2 + n^2 - q^2

How do I show that either a, b, or c is congruent to 0 (mod 5) using this?  Or can I?
..............................................
Cute.  Now that you mention it, I do recall that the P.T.'s seem to have a member == 0 mod 5. (== is the 'congruent' symbol here)  

Such as  3,4,5, or 5,12,13 or  8,15,17 or 7,24,25.  WOW.
.................................................
Now there is a rule that says all P.T.'s can be generated this way:

Pick integers  m,n that are relatively prime: (actually they don't have to be, but it's more fun)

Then:  m^2 - n^2,  2mn, and  m^2 + n^2  are a P.T.
....................................
Easy to show that it works.

(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2

m^4 - 2 m^2 n^2 +  n^4 + 4m^2n^2 = m^4 + 2 m^2 n^2 +  n^4

etc.
...............................
Now suppose  m == p mod 5,  and  n ==  r mod 5
If either  p or r is zero, then 2mn is a multiple of 5, and we are done.

So  m == +-1,+-2  mod 5, and

   n == +-1,+-2  mod 5

We can now try some cases.  (You won't need 64 cases now.)

If m == n mod 5, then m^2 = n^2 mod 5 and  m^2 - n^2 == 0 is the one we want.
If m == -n mod 5, as in  m == -2, n == 2, we again have m^2 - n^2 == 0 mod 5.

So we are left with m == +-1 mod 5 and n == +-2 mod 5, or vice versa.

Now  m^2 == 1 mod 5 and n^2 == 4 mod 5, and then m^2 + n^2 == 0


---------- FOLLOW-UP ----------

QUESTION: I'm still a little confused.  When I choose integers m and n that are relatively prime, how does this relate back to a, b, or c?

Is a==m^2-n^2?  b==2mn?  c==m^2+n^2?

One more thing I forgot to mention when I sent the question, but I only need to work with 0, 1, 2, 3, 4.  However, I need to prove that a==0 mod 5, b==0 mod 5, or c==0 mod 5.

Answer
Questioner:   Jen
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Pythagorean Triple Proof
Question:  QUESTION: Hi I have a question concerning the Pythagorean triple, which mean that there are three natural numbers x, y, z provided that x<y<z.  I have to assume a, b, and c form a Pythagorean triple and prove that either a is congruent to 0 (mod 5), b is congruent to 0 (mod 5), or c is congruent to 0 (mod 5).

If I work this by cases, there will be 64 cases!  I have already worked a couple (ranging from a=5m+1, b=5n+2, and c=5q+3).  I notice that for each I get

5r=m^2 + n^2 - q^2

How do I show that either a, b, or c is congruent to 0 (mod 5) using this?  Or can I?

ANSWER: Questioner:   Jen
Category:  Advanced Math
Private:  No

Subject:  Pythagorean Triple Proof
Question:  Hi I have a question concerning the Pythagorean triple, which mean that there are three natural numbers x, y, z provided that x<y<z.  I have to assume a, b, and c form a Pythagorean triple and prove that either a is congruent to 0 (mod 5), b is congruent to 0 (mod 5), or c is congruent to 0 (mod 5).

If I work this by cases, there will be 64 cases!  I have already worked a couple (ranging from a=5m+1, b=5n+2, and c=5q+3).  I notice that for each I get

5r=m^2 + n^2 - q^2

How do I show that either a, b, or c is congruent to 0 (mod 5) using this?  Or can I?
..............................................
Cute.  Now that you mention it, I do recall that the P.T.'s seem to have a member == 0 mod 5. (== is the 'congruent' symbol here)  

Such as  3,4,5, or 5,12,13 or  8,15,17 or 7,24,25.  WOW.
.................................................
Now there is a rule that says all P.T.'s can be generated this way:

Pick integers  m,n that are relatively prime: (actually they don't have to be, but it's more fun)

Then:  m^2 - n^2,  2mn, and  m^2 + n^2  are a P.T.
....................................
Easy to show that it works.

(m^2 - n^2)^2 + (2mn)^2 = (m^2 + n^2)^2

m^4 - 2 m^2 n^2 +  n^4 + 4m^2n^2 = m^4 + 2 m^2 n^2 +  n^4

etc.
...............................
Now suppose  m == p mod 5,  and  n ==  r mod 5
If either  p or r is zero, then 2mn is a multiple of 5, and we are done.

So  m == +-1,+-2  mod 5, and

  n == +-1,+-2  mod 5

We can now try some cases.  (You won't need 64 cases now.)

If m == n mod 5, then m^2 = n^2 mod 5 and  m^2 - n^2 == 0 is the one we want.
If m == -n mod 5, as in  m == -2, n == 2, we again have m^2 - n^2 == 0 mod 5.

So we are left with m == +-1 mod 5 and n == +-2 mod 5, or vice versa.

Now  m^2 == 1 mod 5 and n^2 == 4 mod 5, and then m^2 + n^2 == 0


---------- FOLLOW-UP ----------

QUESTION: I'm still a little confused.  When I choose integers m and n that are relatively prime,

>> Actually, as I said, they don't really have to be, but you would like your a,b,c to be.

how does this relate back to a, b, or c?

Is a==m^2-n^2?  b==2mn?  c==m^2+n^2?

>>  YES, YES, YES.  Of course, you mean:

Is a = m^2-n^2?  b = 2mn?  c = m^2+n^2?

I was using the == symbol for 'is congruent to'.  (I don't know how to make that 'three-line' thing.)

One more thing I forgot to mention when I sent the question, but I only need to work with 0, 1, 2, 3, 4.  However, I need to prove that a==0 mod 5, b==0 mod 5, or c==0 mod 5.

>> Of course, but:

  saying  x == -1 mos 5 and x == 4 mod 5 are the same.  

And saying  x == -2 mos 5 and x == 3 mod 5 are the same.