# Smallest Right Triangle

Discussion in 'Recreational Math' started by Jim Dars, Dec 16, 2006.

1. ### Jim DarsGuest

Hi All,

Consider a right triangle whose legs, X and Y, and hypotenuse Z are all
integers.
What is the smallest such triangle such that

1) X + Y is a square
2) Z is a square

Not an easy problem. Merry Christmas and

Best wishes, Jim

Jim Dars, Dec 16, 2006

2. ### ProginoskesGuest

Not too hard, though, once you know the formula that generates all
Pythagorean triples:

X = u^2 - v^2
Y = 2 u v
Z = u^2 + v^2,

where u > v > 0. (If you insist further that gcd(u,v) = 1, you get all
"primitive" Pythagorean triples:
gcd (X,Y,Z) = 1.)

Problem (2) is easily answered; Z >= 3^2 + 4^2 = 5^2. Thus u = 4, v =
3, which gives
(X,Y,Z) = (9, 24, 25).

To answer problem (1), you need to find the smallest u and v such that
u^2 + 2 u v - v^2 is a perfect square. If we let this expression equal
K^2, then
2 u^2 - K^2 must be a perfect square. From a Maple search, it appears
that
u = 5 and v = 4 (and hence (X,Y,Z) = (9, 40, 41)) is the solution to
(1).

--- Christopher Heckman

Proginoskes, Dec 17, 2006

3. ### Patrick HamlynGuest

Yah v.good. Now solve it again as if it were one problem with two constraints.

Patrick Hamlyn, Dec 17, 2006
4. ### Jim DarsGuest

Hi Proginoskes,

Not even close. You must understand I posed ONE problem, not two. So X + Y
is a square for the same triangle Z is a square.

Best wishes, Jim

Jim Dars, Dec 17, 2006
5. ### MichaelGuest

Jim Dars, not easy? You can say that again!
This is a problem for the big boys
(Fermat, Euler, Lagrange, etc.).
Fermat first proposed and solved it.
Smallest triplet has sides which are 13-digit integers.
--Michael

Michael, Dec 17, 2006
6. ### ProginoskesGuest

Oh. Well, usually, that's indicated by putting the word AND between the
two conditions. Based on other responses to this thread, that will take
a little more time.

--- Christopher Heckman

Proginoskes, Dec 17, 2006
7. ### Chip EasthamGuest

Extending Christopher's approach to Z being a square as well
as X+Y, we express u = m^2 - n^2 and v = 2mn, & then we ask
X+Y = u^2 + 2uv - v^2 = m^4 + 4m^3n - 6m^2n^2 - 4mn^3 + n^4
to be a square. Setting r = m/n, we can then search for the
rational points on:

y^2 = r^4 + 4r^3 - 6r^2 - 4r + 1 = f(r)

with the proviso that X,Y > 0, which wlog implies u > v > 0:

m^2 - n^2 > 2mn > 0 (and wlog m,n > 0)
(m - n)^2 = m^2 - 2mn + n^2 > 2n^2
m - n > n*sqrt(2)
m > n*(1 + sqrt(2))
r > 1 + sqrt(2) ~ 2.41421...

Note that the curve above has rational points at r = 0, 3/2.
More than likely this makes it tractable to compute more
rational points on the elliptic(?) curve.

regards, chip

Chip Eastham, Dec 18, 2006
8. ### Jim DarsGuest

Hi Chip, Michael, All,

Michael named Fermat quite correctly. Least I give my friends a Black
Holiday present I'll note that he used the process of "Infinite Descent",
but then goes in the reverse direction. Not much of a clue, and it wasn't
meant to be. --- Just concerned that people don't waste time going down
blind (or difficult) alleys.

Best wishes, Paul

Jim Dars, Dec 18, 2006
9. ### MichaelGuest

Chip, I suspect that you may be on the right track or on a right track. A
successive search for rational points by the "Method of Diaphantus" (in
ascent or descent, as used by Fermat, as stated by Jim Dars) *may* be the
trick or a trick.

regards, michael

Michael, Dec 18, 2006
10. ### a1jrjGuest

I gave up quickly with this, although the solution found by Fermat as
noted uses fairly elementary stuff, as in this link...
<http://www.mathpages.com/home/kmath022/kmath022.htm>
HTH
JJ

a1jrj, Dec 18, 2006
11. ### MichaelGuest

Christopher and Chip, there is nothing new below.
I am restating your work, dotting the i's and crossing the t's, mainly for
my own sake.

Given:
(X,Y,Z,W) are positive integers; gcd(X,Y) = 1
X^2 + Y^2 = Z^2 (1)
Z= W^2 (2)
X + Y = W^2 (3)

Find smallest (X,Y)

Let (u,v) be positive integers, one odd, one even; u > v ; gcd(u,v) = 1 :
X = (u^2 - v^2) (odd)
Y = 2uv (even)
Z = (u^2 + v^2) (odd)
Then (X,Y,Z) is a solution of (1).

From (2),
Z = u^2 + v^2 = W^2
Solve for u, v and W.
Let (p,q) be positive integers, one odd, one even; p > q ; gcd(p,q) = 1 :

Then we have two cases:
Case 1: (p^2 - q^2) < 2pq (4)
Case 2: (p^2 - q^2) > 2pq (5)

=========================
Case 1: (p^2 - q^2) < 2pq (4)
=========================

u = (2pq) (even)
v = (p^2 - q^2) > 0 (odd)
W = (p^2 + q^2) (odd)

X = (u^2 - v^2) (odd) = 4p^2q^2 - (p^2 - q^2)^2 > 0
Y = (2uv) (even) = 4pq(p^2 - q^2) > 0
Z = (u^2 + v^2) (odd) = (p^2 + q^2)^2 = W^2

(X + Y) is seen to be of the form 4*k_1 - (4*k_2 + 1) - 4k_3 = ( 4*k -
1) and so cannot be a perfect square.

==========================
Case 2: (p^2 - q^2) > 2pq (5)
==========================

u = (p^2 - q^2) > 0 (odd)
v = (2pq) (even)
W = (p^2 + q^2) (odd)

X = (u^2 - v^2) (odd) = (p^2 - q^2)^2 - 4p^2q^2 > 0
Y = (2uv) (even) = 4pq(p^2 - q^2) > 0
Z = (u^2 + v^2) (odd) = (p^2 + q^2)^2 = W^2

(X + Y) is seen to be of the form (4*k_2 + 1) - 4*k_1 - 4k_3 = ( 4*k
+1) and so can be a perfect square.

(X+Y) = p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 = W^2 (6)
Or, equivalently,
(p^2 - q^2 + 2*pq)^2 - 8*(pq)^2 = W^2 (7)

The conditions are:
p,q positive integers, one odd, one even; p > q ; (p^2 - q^2) > 2pq
(8)

Therefore, for fixed p,
q^2 +2pq - p^2 < 0
(q + p)^2 < 2*p^2
q < (sqrt(2) - 1)*p
q < 0.4142*p (9)

Therefore, in principal, we could write a couple-of- lines program in
Mathematica or Maple to obtain the smallest positive rational integer
solution in a few milliseconds. We compute LHS of (7) for p = 1... N
; q = 1 ..... [0.4142*p] until it is a perfect square.

The real task is to algebraically bootstrap our way up to the smallest
rational integer solution from some starting poin, (which is what Chip has
suggested.)

regards, michael

Michael, Dec 18, 2006
12. ### MichaelGuest

Two questions come to mind:
First, Jim Dars (and Fermat) asked for the smallest solution.
In the reference cited, there is no proof that the solution found is the
smallest one.
Second, the approach gives one the the impression of being a 'hit or miss'
affair and not a systematic one.
--Michael

Michael, Dec 18, 2006
13. ### a1jrjGuest

Excellent analysis - a short Mathcad program took two seconds to find
the smallest solution of p :-
p = 1469 q = 84
with the same result as Fermat in the link. Which demonstrates it is
the smallest solution.

I think there is method in the cited link - basically a re-application
of completing the square until the difference is a linear form which
can be set to zero for a given rational number, but I agree it looks
haphazard rather than systematical.

JJ

a1jrj, Dec 18, 2006
14. ### MichaelGuest

I admit that there is method, which method perhaps falls under the umbrella
of "Method of Diophantus".
In addition to an appearance of being haphazard it does not seem to ensure
that the smallest solution has been found.
--michael

Michael, Dec 18, 2006
15. ### ProginoskesGuest

Yes, but (1) Fermat found the answer systematically, and (2) Fermat
didn't have Mathematica or Maple.

--- Christopher Heckman

Proginoskes, Dec 19, 2006
16. ### MichaelGuest

Christopher, I do not *know* what *systematic* method Fermat used, which
statement does not imply that he did not use a systematic one.
--michael

Michael, Dec 19, 2006
17. ### sttscitransGuest

Yes, Gaussian integers give a quicker derivation

x^2 +y^2 = z^4 =>
(x +iy) = (p+iq)^4 =>
x = p^4 -6p^2q^2 + q^4, y = 4ab(a^2-b^2) =>
p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 =w^2

Assuming p,q, >0
|x| + |y| = u^2 => sgn(x,y)=1 =>

p^4 -6p^2q^2 + q^4 >0 and 4ab(a^2-b^2) >0
or
p^4 -6p^2q^2 + q^4 <0 and 4ab(a^2-b^2)<0

p^4 -6p^2q^2 + q^4 =0 =>

p/q = 1 +sqrt(2), -(1+sqrt(2), (1-sqrt(2), -(1-sqrt(2)

=> p/q > 1+sqrt(2) or sqrt(2)-1 < p/q <1

If sgn(x,y) = -1, then you get solutions to the
problem requiring the hypotenuse to be a square
and the difference of the other two sides to be a square
e.g 119^2 + 120^2 = 169^2

Is there any reason to believe
p^4 + q^4 - 6*p^2*q^2 + 4*p^3*q - 4*p*q^3 =w^2
has an infinitude of integer solutions ?
I have only found 1 solution for each case, although
I have not searched very far.

(p^2 - q^2 + 2*pq)^2 - 8*(pq)^2 = W^2 (7)

This is a form of Pell's equation s^2 -2t^2 = 1
If (p^2 - q^2 + 2*pq)^2 - 2*(2pq)^2 = 1

Is there more than one right-angled triangle with sides
n, n+1, m^2 ?

sttscitrans, Dec 19, 2006
18. ### MichaelGuest

Elegant!
Glad to see you clarify the significance of Case 1.
One of the joys of solving puzzles socially rather than onanistically is the
diversity of perspectives one is exposed to.

I believe that the current wisdom is that algebraic equations (with rational
integer coefficients) of order greater than three, *in general*, tend to
have none or very few rational integer solutions.

This had occurred to me but I dismissed the thoughts, believing this to be a
different and more difficult problem. I will look again for a possible
connection.
--michael

Michael, Dec 19, 2006
19. ### Jim DarsGuest

Hi All,

After a couple of days away I noticed in Michael's initial summation

*********
Given:
(X,Y,Z,W) are positive integers; gcd(X,Y) = 1
X^2 + Y^2 = Z^2 (1)
Z= W^2 (2)
X + Y = W^2 (3)
**********************

Given:
(X,Y,Z,W) are positive integers; gcd(X,Y) = 1
X^2 + Y^2 = Z^2 (1)
Z= W^2 (2)
and not
X + Y = W^2 (3)
but
X + Y = E^2 (3) (assuming someone hasn't introduced E into the
analysis.)??

Best wishes, Jim

Jim Dars, Dec 19, 2006
20. ### MichaelGuest

Yes.
Jim Dars you have a sharp eye!
--michael.

Michael, Dec 19, 2006