Is parametrization necessary?

We wish to construct a Pythagorean triple $a^2 + b^2 = c^2$ with $b = a \pm 1$ .

It is well-known that Pythagorean triples can be generated by choosing integers $p > q > 0$ and setting $a = p^2 - q^2,\ b = 2pq,\ c = p^2 + q^2.$ With $(p,q) = (2,1)$ we get the given triple $(3,4,5)$ . Also, set $r = p - q$ ; then $r^2 = c - b$ .

In order for $b = a \pm 1$ , we must have $(p - q)^2 = 2q^2 \pm 1$ . Rewrite this as $r^2 - 2q^2 = \pm 1$ . This is Pell's equation.

It is know that the positive Pell's equation has infinitely many solutions $(q_n,r_n)$ . The integers $r_n/q_n$ form a sequence with limit $\sqrt 2$ . The first few terms are $\begin{array}{ll} r_n/q_n & a^2 + b^2 = c^2 \\ \hline 3/2 & 21^2 + 20^2 = 29^2 \\ 17/12 & 695^2 + 696^2 = 985^2 \\ 99/70 & 23661^2 + 23660^2 = 33461^2 \end{array}$

Each next term may be generated recursively as $(r',q') = (3r + 4q, 2r + 3q)$ . Since this process can be continued indefinitely, there are infinitely many solutions.

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r = 1
q = 0
repeat
   new_r = 3*r + 4*q
   new_q = 2*r + 3*q
   r = new_r
   q = new_q

   p = r + q
   a = p*p - q*q
   b = 2*p*q
   c = p*p + q*q

   print "sides " + a + ", " + b + "; diagonal: " + c
end repeat

I will solve by reducing the problem to solving Pell's equation. Lets consider the polynomial $f(n) = n^2 + (n+1)^2$ . This polynomial happens to have an interesting property that we will exploit. We wish to solve the equation $f(n) = b^2$ in order to find the side lengths of a square with the properties given in the problem. If we solve for positive $n$ we get that $n = \frac{\sqrt{2b^2 - 1} - 1}{2}$ .

Let's for a moment suppose that $2b^2 - 1$ is a perfect square. $2b^2 - 1$ is an odd number, and therefore its square root will also be odd and if we subtract 1 to this odd number we get an even number. Therefore if $2b^2 - 1$ is a perfect square then $n$ is an integer, and as long as $b>1$ , n will be larger than 0. This means we have no difficulties in studying the discriminant instead of the polynomial. Therefore we now want to solve the equation $2b^2 - 1 = m^2$ to find when this discriminant is a perfect square. Arranging this properly we get $m^2 - 2b^2 = -1$ . This is Pell's equation, for which we may obtain infinitely many solutions.

One subset of rectangles with the given properties has positive integer side lengths of $a$ and $a + 1$ and a positive integer diagonal of $c$ such that it satisfies Pythagorean’s Theorem $a^2 + (a + 1)^2 = c^2$ . This can be rearranged to $(2a + 1)^2 + 1 = 2c^2$ , which is in the form of $x^2 + 1 = 2y^2$ for positive integers $x = 2a + 1$ and $y = c$ .

We can generate Pythagorean triples $(a’, b’, c’)$ by a Pythagorean generator for which $a’ = p^2 - q^2$ , $b’ = 2pq$ , and $c’ = p^2 + q^2$ for positive integers $p$ and $q$ where $p > q > 0$ , but in this case we will also place the given restriction that $b’ = a’ + 1$ , so that $2pq = p^2 - q^2 + 1$ which can be rearranged to $(p - q)^2 + 1 = 2q^2$ , which is also in the form of $x^2 + 1 = 2y^2$ but for positive integers $x = p - q$ and $y = q$ .

Using these similar forms we can set $y = q = c$ and $x = p - q = 2a + 1$ in an attempt to create a recursive equation for $a$ , $b$ , and $c$ . Then $q = c$ and $p = 2a + 1 + q = 2a + 1 + c$ . Substituting these into $a’$ , $b’$ , and $c’$ and simplifying (and using the above equation $(2a + 1)^2 + 1 = 2c^2$ ) gives:

$a’ = 2c^2 + 4ac + 2c - 1$

$b’ = 2c^2 + 4ac + 2c$

$c’ = 3c^2 + 4ac + 2c - 1$

First of all, $a’$ , $b’$ , and $c’$ must be integer sides of a right triangle because they were generated by a Pythagorean generator. Secondly, the above equations for $a’$ and $b’$ show that $b’ = a’ + 1$ . Finally, if $a$ , $b$ , and $c$ are positive integers, then $a’ > a$ , $b’ > b$ , and $c’ > c$ , which ensures that each triplet generated is bigger than the last.

Therefore, the above equations for $a’$ , $b’$ , and $c’$ are recursive equations that can be continued indefinitely. (Using the given example $(a, b, c) = (3, 4, 5)$ as the seed, we can obtain the triplets $(119, 120, 169)$ , $(137903, 137904, 195025)$ , and so on.) There are therefore infinitely many ways to construct a rectangle satisfying the given properties.

Although there is no limiting factor and we can keep generating numbers infinitely fulfilling the given conditions.

But there are only 8 sets in a million and 12 up to a billion numbers. I am not sure whether we should call it "infinitely many ways" OR "finitely many ways". Let us hear from gurus.

Thanks

We are looking for integers $x,z$ such that:

$x^2+(x+1)^2=z^2$

which is equivalent to

$2x^2+2x+(1-z^2 )=0$

Solving for $x$ , we get the positive solution:

$\displaystyle x=\frac{\sqrt{2z^2-1}-1}{2}$

In order to obtain an integer solution, the discriminant of the equation must be a perfect square, i.e. there must exist $t \in \mathbb{Z}$ such that:

$2z^2-1=t^2$

The latter can be rewritten as

$t^2-2z^2=-1$

which is Pell's equation. All the solutions $(t_n,z_n)$ to this equation can be calculated using the relation:

$(1+\sqrt{2})^{2n+1}=t_n+z_n\sqrt{2}$

for $n=1,2,3,...$

As a consequence, all the (infinitely many) Pythagorean triples $(x_n,y_n,z_n)$ , with $y_n=x_n+1$ , can be generated by the formulas ( $n=1,2,3,…$ ):

$\displaystyle x_n=\frac{(1+\sqrt{2})^{2n+1}+(1-\sqrt{2})^{2n+1}}{4}-\frac{1}{2}$

$y_n=x_n+1$

$\displaystyle z_n=\frac{(1+\sqrt{2})^{2n+1}-(1-\sqrt{2})^{2n+1}}{2\sqrt{2}}$

For $n=1,2,…, 20$ we obtain the following:

\( \begin{array}{} x_n & y_n & z_n \\ 3 & 4 & 5 \\ 20 & 21 & 29 \\ 119 & 120 & 169 \\ 696 & 697 & 985 \\ 4059 & 4060 & 5741 \\ 23660 & 23661 & 33461 \\ 137903 & 137904 & 195025 \\ 803760 & 803761 & 1136689 \\ 4684659 & 4684660 & 6625109 \\ 27304196 & 27304197 & 38613965 \\ 159140519 & 159140520 & 225058681 \\ 927538920 & 927538921 & 1311738121 \\ 5406093003 & 5406093004 & 7645370045 \\ 31509019100 & 31509019101 & 44560482149 \\ 183648021599 & 183648021600 & 259717522849 \\ 1070379110496 & 1070379110497 & 1513744654945 \\ 6238626641379 & 6238626641380 & 8822750406821 \\ 36361380737780 & 36361380737781 & 51422757785981 \\ 211929657785303 & 211929657785304 & 299713796309064 \\ \end{array} \)

x² + x² + 1 = y² where all values are integers 2x² + 1 = y² therefore y = sqrt(2x² + 1) for every x there must be some corresponding y value

a^2 + b^2 = c^2 => Every three sets of integers(a,b,c) following this condition are called Pythagorean triplets. I.e. 3^2 +4^2 =5^2 => (3,4,5) .......... We can generate numbers infinitely which satisfy the theorem. So the answer is, "yes, infinitely many ways"