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7/29/2019 The Pythagorean triples whose hypotenuse and the sums of the legs are squares
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Bulletin of Society forMathematical Services and Standards ISSN: 2277-8020Vol. 2 No. 3 (2013), pp. 60-73 www.ijmsea.com
Bulletin of Society forMathematical Services and Standards
60
The Pythagorean triples whose hypotenuse
and the sums of the legs are squares
PROF. DR. K. RAJA RAMA GANDHI1
AND REUVEN TINT2
Resource person in Math for Oxford University Press and Professor at BITS-Vizag1
Number Theorist, Israel2
Abstract. More than 370 years ago the famous French mathematician Pierre de Fermat proposed tosolve the following problem: to find a Pythagorean triples whose hypotenuse and the sum of the legs
were squares, which, despite its simplicity, has been very difficult. Problems associated with its
solution involved many mathematicians such as (Leonhard Euler, Joseph-Louis Lagrange, Ljnggren,
Wacaw Sierpiski and etc.) But in the end it did not reach solution. In our article, solutioncommunicated to obtain the equations giving the required values all elements of the Pythagorean
triples in positive integers (natural) are co-prime integers, and provides a second solution of thisproblem (the values of x, y, z of 45 digits), and some consequences.
1
In 1643, Fermat challenged Mersenne to find a Pythagorean triplet whose hypotenuse and sum of the
legs were squares. (very difficult to solve the following problem, Diophantus of Alexandria,Arithmetica, , 1974 p. 309) Fermat found the smallest such solution.
Problems associated with the solution of this problem, were engaged Euler, Lagrange, Ljunggrenand etc. The first person who show how to obtain first solution of the problem, that resulted by
Fermat was the Wacaw Franciszek Sierpiski. He also solved the problem of obtaining all resultsthat include coprime solutions. ( , , 1959, 12). However, he
has not led the explicit equations for solutions to this problem. Therefore, we believe that it makes
sense to bring below its a solution different from the mentioned, with the corresponding equationsand give the value of the elements of the second triangle of the required type.
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Bulletin of Society forMathematical Services and Standards ISSN: 2277-8020Vol. 2 No. 3 (2013), pp. 60-73 www.ijmsea.com
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2
2.1. Firstly,we make one comment.
Let
are any integer solutions of the equation
, where
Since, for the Pythagorean equation we have
If
Then,
in this case there will always be a rational number.
Comment:[ Ljunggren proved that the equation
has only two solutions in positive integers: 1 and 13 (., 250
, , 1968 p. 146). Lagrange belongs recurrence relation, throughwhich can be found all solutions of the equation
in rational numbers (. , , ,
, 1961, p. 80).]
Therefore, the challenge is to find all the integer values are coprime and , which are
then given and (become known value and ) allow to obtain the required values
of all uniquely and in a positive integer (natural) are coprime numbers.
2.2. In the Pythagorean triple
Let );(2;111
2
1
2
1nmmnnmm
then we obtained using the identity:
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Bulletin of Society forMathematical Services and Standards ISSN: 2277-8020Vol. 2 No. 3 (2013), pp. 60-73 www.ijmsea.com
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, we have
and we obtain one of the variants of the recurrence methods for all solutions of the equations
and
in the rational coprime integers:
(If and are fractional numbers, then equal to the numerator of the fraction as described
numerator, but equal to the numerator of the fraction as the denominator described);
( with the initial values ),such that, all integers that are coprimevalues and :Comment:
[The recurrence method of Lagrange could be obtained in a simple substitution value in
only using other notation
- The recurrence method of Lagrange for solution of equation
in rational numbers.]
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The values given at the beginning of our article.
and etc.2.3. For all coprime solutions of the problem in natural numbers is a necessary and sufficient
condition, that in
(Hypotenuse trivial could be less than the sum of the two legs).
2.4. We have in and choosing the lowest absolute value of two of its corresponding
from and , we finally obtain the following using the method:
,where and etc.
Comment:[For two integers and we get the maximum value from two by absolute
value .]
It follows that in
2.4.1. Indeed,
and
, since otherwise there will be , if is rational then is impossible.
2.4.2. It must be for the same reasons
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For example,
2.4.3. For from two
one of them must be
and
To conclude the proof, it remains to note with respect to that recurrence relations is unbounded
, this means that we obtain an infinite set of Pythagorean triples whose hypotenuse and the
sums of the legs are squares, and with respect to and the sequence gives all the solutions of
the problem. Arguing as above, this completes the proof totally.
Comment:
[In the relation does not depend on one, but two parameters , included
in this equation. And if is a necessary and sufficient condition for the unique
determination of required values , then for the next value this
condition is only necessary, but with
would be a necessary and sufficient.]
2.5. The second solution of the problem in natural numbers corresponding to which is definedbelow:
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3
It follows that we have identities:
for arbitrary and .
With initial values , or as defined above identities are transformed intorecurrence equations for all rational solutions of the equations
, as well as all solutions in positive integers equations
4
Assume that it makes sense to give these two analogies of Fermat's problem:
4.1.
Using the same notation and procedure as in the previous case, we come to one embodiment of the
recurrence methods for solutions of the equation
in the rational numbers:
for
Similarly,
and
and, therefore, the integer values and .
For
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and etc.
4.2. The second analogy of Fermat's problem:
To solve the problem could be to use the previous method, but we use the other. If
,then
For
References:
1. Diophantus of Alexandria Arithmetica , , - , , 1974.
2. W. Sierpiski, , - , , 1959.
3. W. Sierpiski, 250 , , ,1968.
4. W. Sierpiski, , , 1961.
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,
-
PROF. DR. K. RAJA RAMA GANDHI1
AND REUVEN TINT2
Resource person in Math for Oxford University Press and Professor at BITS-Vizag1
Number Theorist, Israel2
. 370 .
: ,
-, , ,
. , , (,, , .) .
, ()
, ( 45 ) .
.
1
. 1643 (
,, ,, 1974 . 309)
.
, , , , . , , . .
( , , 1959, 12).
. ,
, , .
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2
2.1. .
-
,
,
.:
[ ,
: 1 13 (., 250
, , 1968 . 146).
,
(. , , ,, 1961, . 80).]
, ( )
() .
2.2.
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);(2;111
2
1
2
1nmmnnmm
, :
,
,
:
( , ,
);
( ),, ,
::
[
-
.]
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.
..2.3.
,
( ). 2.4.
, :
, ..
:
[
.]
2.4.1.
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, , .
2.4.2. ,
,
2.4.3.
,
,, , ,
-,
. .
:
[ , ,
.
,
,
.]
2.5. , , :
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3
:
.
,
,
4
, : 4.1.
, ,
:
,
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, , .
..4.2. :
,
.
,
:
1. , , - , , 1974.
2. . , , - , , 1959.
3. . , 250 ,,, 1968.
4. . , , , 1961.