How do you generally prove that the sum of a right triangle's hypotenuse and it's base have a common factor with the perpendicular side?

Question

I recently came across this one observation in my research on right triangles:

In any Pythagorean triangle, the sum of the hypotenuse and the base is a multiple of, at least one factor of the base, other than 1.

eg: Let $a = 28$, $b = 45$, $c = 53$.

Then $45 + 53 = 98$, with who $28$ shares a common factor: $7$.

I have been able to prove this when $a = 2m$, $b = m^2 - 1$, $c = m^2 + 1$ AND when $a = x^{1/2}$, $b = (x-1)/2$, and $c = (x + 1)/2$

I have tried and tested the same for many triangles which do not follow the above mentioned criterion, and they seem to work fine, like the example given above. I tried to use Dickson's method of taking $a = r + s$, $b = r + t$, and $c = r + s + t$, as it would exhaust all triples. It did not work out fine. Does anybody have any other method on how to go about it?

Thanks,

S Sandeep.

Answer 1

You don't need a production scheme for all pythagorean triples to solves this problem.

From $a^2=c^2-b^2=(c+b)(c-b)$ it follows that any prime factor of $c+b>1$ must occur in $a$.

Answer 2

The following will generate all Pythagorean triples uniquely $$a=k(m^2-n^2)=k(m-n)(m+n),\ \ \ b=k\cdot(2mn),\ \ \ c=k(m^2+n^2)$$ where $m, n$, and $k$ are positive integers with $m \gt n, m − n$ odd, and with $m$ and $n$ coprime such that $a^2+b^2=c^2$. (For example, $(k,m,n,a,b,c)=(1,7,2,45,28,53)$ is the case you wrote.)

Case 1 : $c+b=k(m^2+n^2)+k\cdot(2mn)=k\cdot (m+n)^2$ and $a=k(m-n)(m+n)$ have a common factor $k(m+n)\gt 1$ (even if $k=1$).

Case 2 : $c+a=k(m^2+n^2)+k(m^2-n^2)=2m^2k$ and $b=2mnk$ have a common factor $2mk\gt 1$ (even if $k=1$).

Answer 3

Since $a, b, c$ share a common factor, then the result becomes trivial. Therefore, let's consider the case where $a, b, c$ do not have a common factor.

$a^2 + b^2 = c^2 \implies a^2 = c^2 - b^2 = (c+b)(c-b)$

Now, by fundamental theorem of arithmetic, let $a =\prod\limits_{p_i \in \text{Primes}} p_i^{a_i} \implies a^2 = \prod\limits_{p_i \in \text{Primes}} p_i^{2 a_i}$.

Think of above statement as $a$ being written as product of its prime factors, like $a = p_1 \cdot p_2^3 \cdot p_3 \cdots$.
$a^2$ will be the same expression, squared.

Therefore, $(c+b)(c-b) = a^2 = \prod\limits_{p_i \in \text{Primes}} p_i^{2 a_i}$.

Therefore, each prime $p_i$ in the prime factorization of $a$ has to appear at least twice in the product $(c+b)(c-b)$.
Now, $(c+b) > (c-b) \ge 1$. This is because least possible value of $c-b$ is $1$. So, $c+b$ contributes at least one prime $p_i$ to the overall product (possibly more, or even all when $c-b = 1$).

But $p_i$ also present in $a$. Therefore, $c+b$ shares at least one factor with $a$.