**Palindromic Pythagorean Triples** are defined and calculated by this extraordinary intricate and excruciatingly complex formula.

So, this line is for experts only _{}

pal_X^{2}+ pal_Y^{2}= pal_Z^{2}

Here are a few general websources that will teach you how to construct pythagorean triples.

Pythagorean Theorem

Pythagorean triple - Eric W. Weisstein

Finding **Palindromic** Pythagorean Triples is quite another matter.

I'm counting on you to find them !

[ *January 23, 2005* ]

Zakir Seidov contributed a list of palindromic **areas** of **primitive pythagorean triangles**.

I have displayed them in a separate wonplate. Please find next the reference address ¬

won163.htm

Of course, the well known triplet (3,4,5) really was not too hard to find, was it !

Neither was one of its multiples with

3^{2}+ 4^{2}= 5^{2}

303^{2}+ 404^{2}= 505^{2}

John Abreu sent me a method to find more palindromic multiples of the above pythagorean triplet (3,4,5).

Let S be any finite sequence of 0's and1's, let S' be its reflection; for example, if S=101011, then S'=110101.

Define X:= SS', then X is a palindrome containing only0's and1's

( in the example X :=101011110101).

Besides, if A is any digit from 0 through 9, the number AX := A.X is also a palindrome.

Now, multiply the identity 3^2 + 4^2 = 5^2 by X^2 to obtain:(3X)^2 + (4X)^2 = (5X)^2 The last identity is telling us that the palindromes 3X, 4X and 5X form a pythagorean triple.

The real challenge is to find triplets where the palindromic numbers have NO COMMON factors.

For the moment I can only supply *NEAR MISSES* !

20^{2}+ 99^{2}= 101^{2}

252^{2}+ 275^{2}= 373^{2}

363^{2}+ 484^{2}= 605^{2}

[ *August 29, 1998* ]

Jud McCranie wrote me the following email ¬

" One of your pages asks for Pythagorean Triples that are palindromes, other than ones based on 3-4-5.

I did a search and I didn't find any. Here are ones where both legs arepalindromes. In several cases,

the third number is a "near miss" - changing one digit by one, or exchanging two consecutive digits would

result in apalindrome."3 4 5 6 8 10 363 484 605 464 777 905 3993 6776 7865 6776 23232 24200 313 48984 48985 8228 69696 70180 30603 40804 51005 close 34743 42824 55145 close 29192 60006 66730 25652 55755 61373 52625 80808 96433 36663 616616 617705 48984 886688 888040 575575 2152512 2228137 6336 2509052 2509060 2327232 4728274 5269970 3006003 4008004 5010005 close 3458543 4228224 5462545 close 80308 5578755 5579333 2532352 5853585 6377873 5679765 23711732 24382493 4454544 29055092 29394580 677776 237282732 237283700 300060003 400080004 500100005 close 304070403 402080204 504110405 close 276626672 458515854 535498930 341484143 420282024 541524145 close 345696543 422282224 545736545 close 359575953 401141104 538710545 277373772 694808496 748127700 635191536 2566776652 2644203220 6521771256 29986068992 30687095560 21757175712 48337273384 53008175720 30000600003 40000800004 50001000005 close 30441814403 40220802204 50442214405 close 34104840143 42002820024 54105240145 close 27280108272 55873637855 62177710753

John Abreu (email) gives a method to find more palindromic multiples of the triplet (3,4,5) - go to topic

Jud McCranie (email) made an attempt to find palindromic triples - go to topic

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