

A068536


Numbers m such that m^2 + (reversal of m)^2 is a square. (Leading 0's are ignored.)


5



88209, 90288, 125928, 196020, 368280, 829521, 1978020, 2328480, 5513508, 8053155, 19798020, 86531940, 197998020, 554344560, 556326540, 1960396020, 1979998020, 5543944560, 5925169800, 8820988209, 9028890288, 12592925928, 14011538112, 19602196020, 19799998020
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OFFSET

1,1


COMMENTS

The sequence is infinite, even if it is restricted to terms that end with a nonzero digit, as any term generates an infinite number of other terms by the following scheme: If m is a term of the sequence and d(m) denotes the number of digits of m, then set m' = m*10^d'+m with d' >= d(m). For d' >= d(m) we have reverse(m') = reverse(m)*10^d' + reverse(m) and thus (m')^2 + reverse(m')^2 = (m*10^d'+m)^2 + (reverse(m)*10^d'+reverse(m))^2 = (m^2+reverse(m)^2)*(10^d'+1)^2. As m^2+reverse(m)^2 is a perfect square by assumption, the product on the righthand side of the equation is also a perfect square and m' is part of the sequence. The calculation works also with m' = m*(10^(k*d')+...+10^d'+1). As an example take a(1)=88209. All numbers 8820988209, 882098820988209, 88209882098820988209, ... and 88209088209, 882090088209, 8820900088209, ... are also terms of the sequence.  Matthias Baur, May 01 2020


LINKS

Sheng Jiang and RuiChen Chen, Digits reversed Pythagorean triples, International Journal of Mathematical Education in Science and Technology, volume 29, number 5, 1998, pages 689696.


EXAMPLE

88209^2 + 90288^2 = 126225^2, so 88209 belongs to the sequence.


MATHEMATICA

Do[If[IntegerQ[Sqrt[n^2 + FromDigits[Reverse[IntegerDigits[n]]]^2]], Print[n]], {n, 1, 10^6}]


CROSSREFS



KEYWORD

base,nonn


AUTHOR



EXTENSIONS



STATUS

approved



