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.

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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**