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PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

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PEN Math problems from the collection "Problems in Elementary Number Theory"

Math problems from the collection "Problems in Elementary Number Theory"

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Peter

#1 h May 25, 2007, 12:25 AM • 1 Y

Y by Adventure10

Show that the number of integral-sided right triangles whose ratio of area to semi-perimeter is $p^{m}$ , where $p$ is a prime and $m$ is an integer, is $m+1$ if $p=2$ and $2m+1$ if $p \neq 2$ .

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maxal

#2 h Oct 3, 2008, 10:41 AM • 1 Y

Y by Adventure10

The problem is equivalent to finding the number of solutions in positive integers to the following system of equations:
$\begin{cases} a^2 + b^2 = c^2 \\ ab = (a + b + c)p^m\end{cases}$
Denoting $a + b = t$ , we have $t^2 = c^2 + 2(t + c)p^m$ implying that $(t + c)(t - c) = 2(t + c)p^m$ and
$c = t - 2p^m.$
Therefore, the original system is equivalent to
$\begin{cases} a + b = t \\ ab = 2(t - p^m)p^m\end{cases}$
implying that $a, b$ are the roots of the following quadratic equation:
$x^2 - t x + 2(t - p^m)p^m = 0$
which is solvable in integers only if its discriminant $t^2 - 8 (t - p^m)p^m = (t - 4p^m)^2 - 8p^{2m}$ is a square.

If $(t - 4p^m)^2 - 8p^{2m} = q^2$ then $8p^{2m} = (t - 4p^m - q)((t - 4p^m + q)$ . All possible non-negative such $q$ are obtained as $q = \frac {d_2 - d_1}{2}$ where $d_1 d_2 = 8p^{2m}$ and $d_2 \geq d_1$ are of the same oddness.

If $p = 2$ then $8p^{2m} = 2^{2m + 3}$ and $q = \frac {2^k - 2^{2m + 3 - k}}{2} = 2^{k - 1} - 2^{2m + 2 - k}$ where $k = m + 2\dots,2m + 2$ . These $m + 1$ distinct values of $k$ give the following solutions:
$\begin{cases} a = 2^{k - 1} + 2^{m + 1} \\ b = 2^{2m + 2 - k} + 2^{m + 1} \\ c = 2^{k - 1} + 2^{2m + 2 - k} + 2^{m + 1}\end{cases}$
If $p > 2$ then $q = \frac {4p^k - 2 p^{2m - k}}{2} = 2 p^k - p^{2m - k}$ (for $k = m,\dots,2m$ ) or $q = \frac {2p^k - 4 p^{2m - k}}{2} = p^k - 2 p^{2m - k}$ (for $k = m + 1,\dots,2m$ ). Therefore, the number of solutions here is $2m + 1$ . Explicit solutions can be derived as above.

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#3 h May 24, 2013, 8:29 PM • 1 Y

Y by Adventure10

There is a MUCH simpler proof knowing that all Pythagorean triples can be parametrized as (k(m^2-n^2), 2mnk, k(m^2+n^2) where m, n, and k are positive integers such that m - n is odd and gcd(m, n) = 1. I won't write it here because I don't know how to use Latex and it will look horrible, but that idea just about solves the problem instantly.

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