# How to show if $m$ and $n$ are coprime and $m-n$ is odd, then $(m^2-n^2)^2$, $(2mn)^2$, $(m^2+n^2)^2$ have a common factor?

We know that the Pythagorean triples can be generated by the Euclid's formula $a=m^2-n^2$, $b=2mn$, $c=m^2+n^2$ for any positive integers $m,n$ and $m>n$.

I am trying to prove the statement:

The triple generated by Euclid's formula is primitive if and only if $m$ and $n$ are coprime and $m-n$ is odd.

I could prove the $\Rightarrow$ direction. Now I am trying to show the $\Leftarrow$ proof.

I am trying to prove by contradiction. If $m$ and $n$ are coprime and $m-n$ is odd, let's assume the triple generated is not primitive. Then there is a factor $k$ in each term. But since $m$ and $n$ are coprime and $m-n$ is odd, we can deduce that $(m^2-n^2)^2$ is odd, $(2mn)^2$ is even and $(m^2+n^2)^2$ is odd. Then $k$ can only be odd. Let $k=2p+1$ where $p$ is an integer. But then I am not sure how to proceed to obtain a contradiction.

Any helps are greatly appreciated. Many thanks!

## 3 Answers

Informal outline Suppose for example that $(m^2-n^2)^2$ and $(2mn)^2$ are not relatively prime. Then there is a prime $p$ that divides $m^2-n^2$ and $2mn$. Since $m$ and $n$ have opposite parity, $p\ne 2$.

Since $p$ divides $2mn$, and $p\ne 2$, it follows that $p$ divides one of $m$ or $n$, say $m$. Since $p$ divides $m^2-n^2$, it follows that $p$ divides $n^2$, and hence $n$. This contradicts the fact that $m$ and $n$ are relatively prime.

Remark: We started, as you did, from the assumption that the numbers $(m^2-n^2)^2$, $(2mn)^2$, and $(m^2+n^2)^2$ are not pairwise relatively prime. However, a triple $(x,y,z)$ such that $x^2+y^2=z^2$ is usually called primitive if $x$, $y$, and $z$ are pairwise relatively prime. It makes no real difference, since $x^2$, $y^2$, and $z^2$ are pairwise relatively prime if and only if $x$, $y$, and $z$ are.

• In your remark, I believe you meant $(m^2-n^2)^2$. – Hayden May 20 '14 at 4:37
• Thanks Andre. But just a question, according to mathworld.wolfram.com/PrimitivePythagoreanTriple.html, primitive triples are where gcd(a,b,c)=1 right? So they are not necessarily have to be pairwise coprime? For example gcd(2,4,5)=1, but they are not pairwise coprime. Which one is correct? Thanks. – user71346 May 20 '14 at 4:39
• Thanks for spotting it. Yes, that is what was meant. – André Nicolas May 20 '14 at 4:40
• @user71346: Both are correct, and equivalent. If they are pairwise relatively prime, then $\gcd(a,b,c)=1$. Conversely, if $\gcd(a,b,c)=1$, then they are pairwise relatively prime. Suppose for example to the contrary that $\gcd(a,b)\gt 1$. Then some prime $p$ divides $a$ and $b$. From $a^2+b^2=c^2$ we conclude that $p$ divides $c$, contradicting the fact $\gcd(a,b,c)=1$. In principle the $\gcd(a,b,c)=1$ is a bit better, since it is "weaker." However, the proof of equivalence is so simple that it really doesn't matter. – André Nicolas May 20 '14 at 4:45
• If we start from weaker assumptions, we get an at least apparently stronger theorem, although in this case the two assumptions are so clearly equivalent that it makes no practical difference. I guess one can imagine a situation, in a domain that doesn't have unique factorization, where the exact assumptions could make a difference. – André Nicolas May 20 '14 at 4:59

Suppose $k | (m^2-n^2)$ and $k | (m^2+n^2)$, then $k | 2m^2$ and $k | 2n^2$. But then $k$ divides $\gcd(2m^2,2n^2)=2\gcd(m^2,n^2)=2$. The last step follows from the fact that $\gcd(m,n)=1$. Thus $k=1$ or $k=2$. Ask yourself why can't $k$ be $2$?

• Yes, you are right, if k=2 then it is not primitive. Thanks. – user71346 May 20 '14 at 4:45

I thought I had a proof that, if $$m,n$$ are mutually prime, $$m,n$$ would generate only primitive Pythagorean triples. The $$proof$$ is between the asterisk below. $$\text{*************************}$$

$$\text{We are given }\quad A=m^2n^2\quad B=2mn\quad C=m^2+n^2$$

Let $$x$$ be the GCD of $$m,n$$ and let $$p$$ and $$q$$ be the cofactors of $$m$$ and $$n$$ respectively. Then we have

$$A=(xp)^2-(xq)^2\quad B=2xpxq\quad C=(xp)^2+(xq)^2$$ $$A=x^2(p^2-q^2)\quad B=2x^2(pq)\quad C=x^2(p^2+q^2)$$

If $$GCD(m,n)=1$$, then $$GCD(A,B,C)=1$$ and $$(A,B,C)$$ is a primitive triple. This means that $$m$$ and $$n$$ must be co-prime to generate a primitive.

$$\text{*************************}$$

However, a counter-example destroys to so-called proof: $$\quad\text{Let }m,n=7,3$$. $$A=49-9=40\quad B=2*7*3=42\quad C=49+9=58\quad GCD(40,42,58)=2$$ $$\therefore GCD(m,n)=1\neg\implies GCD(A,B,C)=1$$ I believe but do not know that the proof may be valid if we insist that $$m$$ and $$n$$ be of opposite parity.

The only two formulas I know about that will generate only primitive triplets do not generate all of them but $$C-B=1$$ in the first one and $$C-A=2$$ in the second one.

$$A=2n^2+1\quad B=2n^2+2n\quad C=2n^2+2n+1$$ $$A=4n^2-1\quad B=4n\quad C=4n^2+1$$

$$\mathbf{UPDATE:}$$ I did some looking and found a variation of Euclid's formula that can let us prove when we have found a primitive with more confidence.

$$\text{Let}\quad A=(2m-1+n)^2-n^2\quad B=2(2m-1+n)n\quad C=(2m-1+n)^2+n^2\quad m,n\in\mathbb{N}$$

This formula generates only and all triples where $$GCD(A,B,C)=(2k-1)^2,k\in\mathbb{N}$$ which includes all primitives.

If we let $$GCD((2m-1),n)=x$$ and $$p,q$$ be the respective cofactors, then we have $$A=(xp+xq)^2-(xq)^2\quad B=2(xp+xq)xq\quad C=(xp+xq)^2+(xq)^2$$ $$A=x^2(p+q)^2-x^2q^2\quad B=2x^2(p+q)q\quad C=x^2(p+q)^2+x^2q^2$$ We can see by inspection that if $$(2m-1)$$ and $$n$$ have common factors, then GCD(A,B,C) an odd square and, if GCD((2m-1),n)=1, the triple is primitive.