Product within Pythagorean Triplets

• If $n \mid abc$ for every primitive pythagorean triplets $(a,b,c)$ , then $n \mid abc$ for every primitive/non-primitive pythagorean triplets $(a,b,c)$ . So, we concern ourselves with only the primitive ones.

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• For $(a,b,c)$ to be primitive, exactly one of $a$ and $b$ is even. We'll assume $b$ is the even one, over this discussion.

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• Exactly one of $a$ and $b$ is divisible by $3$ .

Proof : For the sake of contradiction, let both $a^2$ and $b^2$ be $1(\mod 3)$ . Then $c^2=a^2+b^2 \equiv 2 (\mod 3)$ , which is impossible, beacuse no square can be congruent to $2 (\mod 3)$ . To see why, every integer $x$ is $3n$ or $3n \pm 1$ . So, every square $x^2$ must be $0$ or $1$ modulo $3$ . So, at least one of $a$ and $b$ must be divisible by $3$ .

But to be primitive, not both of $a$ and $b$ can be divisible by $3$ . Hence, exactly one of $a$ and $b$ is divisible by $3$ . $\blacksquare$

So, $3 \mid ab \implies 3 \mid abc$ .

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• $(a,b,c)$ is a primitive pythagorean triplet only if $4 \mid b$ .

Proof : As $(a,b,c)=(n^2-m^2,2nm,n^2+m^2)$ gives all the primitive triples and $(n^2-m^2,2nm,n^2+m^2)$ is primitive only when $m$ and $n$ are coprime and of opposite parity, $4 \mid b=2nm$ . $\blacksquare$

So, $4 \mid b \implies 4 \mid abc$ .

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• $5$ divides at least one of $a,b$ and $c$ .

Proof : Every square is $0, 1$ and $4$ modulo $5$ . It's enough to prove that $5 \not \mid a^2$ and $5 \not \mid b^2 \implies 5 \mid c^2 \implies 5 \mid c$ .

In modulo $5$ , $5 \not \mid a^2$ and $5 \not \mid b^2 \implies (a^2,b^2)\equiv (1,1)$ or $(a^2,b^2)\equiv (4,4)$ or $(a^2,b^2)\equiv (1,4)$ or or $(a^2,b^2)\equiv (4,1)$ .

But $(a^2,b^2)\equiv (1,1) \implies c^2=a^2+b^2 \equiv 2$ , which cannot be. For similar reason, we discard $(a^2,b^2)\equiv (4,4)$ .

And both of remaining $(a^2,b^2)\equiv (1,4)$ and $(a^2,b^2)\equiv (4,1)$ implies $c^2 \equiv 0$ modulo $5$ . So, $5 \mid c^2 \implies 5 \mid c$ . $\blacksquare$

Hence, $5 | abc$ .

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• As $3 | abc$ , $4 | abc$ and $5 | abc$ , we have $lcm(3,4,5)=60 \mid abc$ .

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Now, we have, for every primitive/non-primitive triplets $(a,b,c)$ , $60 \mid abc$ ; and as for the triplet $3,4,5$ , we have $abc=60$ , we're assured that $\boxed{60}$ is the largest positive integer that divides $abc$ for every triplets $(a,b,c)$ .

Let $m,n$ be positive integers.Every Pythagorean triplets can be expressed as

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

We need to find the maximum value of n which $n | 2 m n(m^2+n^2)(m^2-n^2)=2 mn (m^4-n^4)$

Recall that $x^4 \equiv 0,1 \pmod{3}$ and $x^4 \equiv 0,1 \pmod{5}$

First deal with the $m n (m^4-n^4)$

Let $m=5k$ and $n=5k+r$ where $1 \leq r \leq 4$ .Now $m^4 \equiv 0 \pmod{5}$ and $n^4 \equiv 1 \pmod{5}$ .So , $(m^4-n^4)$ is not divisible by $5$ .But $m=5k$ and it is multiplied with them.So, overall it is divisible by $5$ .

Again $m=5k+r$ and $n=5k$ .The case is same above.

The last case is $5 |m$ and $5 |n$ then $m^4 \equiv n^4 \equiv 0 \pmod{5}$ .Here $(m^4-n^4)$ is divisible by 5.

If $m=5k_1 +r_1$ and $n=5k_2 +r_2$ then $m^4 ,n^4 \equiv 1 \pmod{5}$ . so, $m^4-n^4 \equiv 0 \pmod{5}$

This logic will be applicable for divisibility by $3$ .

Now $m n (m^4-n^4)$ is divisible by $5 \times 3=15$

If $m$ is odd $n$ is even then $m n (m^4-n^4)$ will be even and vice-versa is also true. If $m,n$ both odd or both even then again it has a at least one factor of two.

So, $m n (m^4-n^4)$ is now divisible by $15 \times 2=30$

Therefore, the whole expression $2 m n (m^4-n^4)$ is always divisible by $30 \times 2=60$