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?
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
Puzzle Resolution
