Difference of two squares relationship

Question

When do we have for $b|c$ that the following relationship holds where $a\neq c$? (all the variables are integers)

$$b^2-1=a^2-c^2$$

Setting $c=2b$ and looking at the resulting Pell's equation quickly gives $$17^2-1=38^2-34^2.$$ It is possible that smaller counterexamples exist. — Jyrki Lahtonen, May 09 '14 at 05:13
@Jyrki Lahtonen Thank you for the catch; but here is an additional piece of information. Python suggest that this statement is indeed true. Any ideas for a proof? — ruadath, May 09 '14 at 06:07
What exactly is the question? What is the claim that Python supports? — Jyrki Lahtonen, May 09 '14 at 06:35
@Jyrki Lahtonen Sorry, moved to different question since it became too different. Please take a look here http://math.stackexchange.com/questions/787537/perfect-square-relationship-with-no-solutions — ruadath, May 09 '14 at 06:36

score 1 · Accepted Answer · answered May 09 '14 at 05:04

Of course for the equation $X^2+Y^2=Z^2+t$

There is a particular solution:

$X=1\pm{b}$

$Y=\frac{(b^2-t\pm{2b})}{2}$

$Z=\frac{(b^2+2-t\pm{2b})}{2}$

But interessuet is another solution: $X^2+Y^2=Z^2+1$

If you use the solution of Pell's equation: $p^2-2s^2=\pm1$

Making formula has the form:

$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$

$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$

$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$

number $L$ and any given us.

The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple. $a^2+b^2=c^2$

This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$

Allows you to find Pythagorean triples with a given difference.

$a=2s(p+s)$

$b=p(p+2s)$

$c=p^2+2ps+2s^2$

$b-a=\pm{k}$

Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.

1 Answers1