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Calling all pythagoreans (Posted on 2006-07-03) Difficulty: 3 of 5
The triangle with sides 3, 4, and 5, is the smallest integer sided pythagorean triangle. Can you prove that in every such triangle:
  • at least one of its sides must be multiple of 3?
  • at least one of its sides must be multiple of 4?
  • at least one of its sides must be multiple of 5?

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Solution Puzzle Resolution | Comment 2 of 7 |

Let  us denote the sides of a Pythagorean triangle as (a,b,c) satisfying  a^2+b^2 = c^2.


CASE 1: One of the numbers (a, b) is divisible by 3

.Suppose not; that is suppose that a,b are divisible by 3.

 Consequently, a^2=1 Mod (3) and b^2 = 1 (Mod 3) ; since 2 is not a quadratic residue in Mod 3 system
This implies that a^2+b^2 = 2 (Mod 3), giving c^2 = 2 Mod (3).
This is not feasible, c^2 must be congruaent to either 0 or 1 (Mod 3).
Hence, at least one of the numbers (a,b)  is divisible by 3, whenever (a,b) = (0,1) Mod 3, (1,0) Mod 3.


CASE 2: One of the numbers (a,b) is divisible by 4

If possible let both  a and b be odd.

Then, a^2 = 1 Mod 4 , and  b^2 = 1 Mod 4; giving c^2 = 2 Mod 4, which is an impossibility, as :

Square of any number must necessarily be congruent to 0 or 1 Mod 4.

Hence, at least one of a and b is even.

WLOG, let us assume that  a is even and  further let us  suppose that  4 does not divide a.

Accordingly, a = 2 Mod 4, giving: a^2 = 4  Mod 8, and

b^2 = 1 Mod 8, as b is odd.

This gives, c^2 = (1+4) Mod 8 = 5 Mod 8, but since 5 is not a quadratic residue in the Mod 8  system, this is not feasible. This is a contradiction., so that if a is even then it must be divisible by 4..

Consequently,   at least one of a and b must be divisible by 4.


CASE 3: One of the numbers (a, b, c) is divisible by 5.

We know that all perfect squares must correspond to 0, 1 or 4 (Mod 5).

If possible, let us suppose that both a and b are not divisible by 5.

Thn, a^2 = 1,4 (Mod 5) and b^2 = 1,4 (Mod 5)

Accordingly, c^2 =a^2+b^2 = 2, 0 or 3 (Mod 5)

Now, c^2 cannot be congruent to 2 or 3 (Mod 5), since the elements 2 and 3 does not correspond to quadratic residues in Mod 5.

Consequently, c^2 = 0 (Mod 5); giving, c = 0 (Mod 5).

Hence, if  a and b are both indivisible by 5, then c is divisible by 5.

 

----------------------------Q E D -----------------------------------------------

 

 

As a corollary, since  each of (3,4) , (4,5) and (3,5) are pairwise coprime, it  follows that,::

a*b*c  is divisible by 3*4*5 = 60.   

 

 

Edited on July 3, 2006, 1:39 pm

Edited on July 3, 2006, 1:40 pm

Edited on July 3, 2006, 1:41 pm

Edited on July 3, 2006, 1:45 pm

Edited on July 3, 2006, 10:32 pm
  Posted by K Sengupta on 2006-07-03 13:36:04

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