Theorem FLT

Pythagorean triples

First, secondary & n-th degree.

Solving with the consistent difference equal degrees integers

Proof of the impossibility of solving Equation FERMAT

MMaannttzzaakkoouurraass DD..NNiikkooss

AAbbssttrraacctt:: IItt iiss wweellll--kknnoowwnn tthhaatt tthheerree aarree mmaannyy ssoolluuttiioonnss iinn iinntteeggeerrss ttoo 222zyx ,, ffoorr iinnssttaannccee

((33,, 44,, 55)),,((55,, 1122,, 1133)).. TThhee BBaabbyylloonniiaannss wweerree aawwaarree ooff tthhee ssoolluuttiioonn ((44996611,, 66448800,, 88116611)) aass eeaarrllyy aass

aarroouunndd 11550000 BB..CC.. AArroouunndd 11663377 [[11]],,PPiieerrrree ddee FFeerrmmaatt wwrroottee aa nnoottee iinn tthhee mmaarrggiinn ooff hhiiss ccooppyy ooff

DDiioopphhaannttuuss’’ AArriitthhmmeettiiccaa ssttaattiinngg tthhaatt nnn

zyx hhaass nnoo ssoolluuttiioonnss iinn ppoossiittiivvee iinntteeggeerrss iiff nn >> 22

..BBuutt rrhhee uussee ooff eeqquuaattiioonn 01)(

mm and finding the λ, that which arises from the

initial nnnzyx according to the method << Solving with the consistent difference equal

degrees integers >> helps effectively solution to the problem.

αα.. PPrriimmaarryy PPyytthhaaggoorreeaann ttrriipplleett iiss aa lliisstt ooff tthhrreeee iinntteeggeerrss ((xx,, yy,, zz)) ssuucchh tthhaatt iinn ggeenneerraall

a x b y c z with , ,a b c Z (1). Such a triad called primary if the integers are non-negative

gcd(x,y,z)=1. There is a simple method for producing Pythagorean triplets degree n <= 2 but is

also the same method of proof of the impossibility for n> 2. This method is called method

consistent difference equal degrees integers, ie integers which differ by an integer constant or

explicit general. But first to apply this method we divide the equation (1) with y, where 0y

and get x z

a b cy y

.

consider my

x and m

y

z where Qm , then substituting in (1) will be

mcbma )( but ca

bam

and

ca

bсm

therefore selectingς )( caky

, Zk implies that primary Pythagorean triad[2] will become

kbaz

kcay

kbcx

)(

)(

)(

Cyclically If we divide the variables {y, z} with x, {x, z} with y, and {x, y} with z shows the

general solution of (1) with the three relations ...

kb)λ(az

kc)(ay

kb)λ(cx

ή

ka)λ(bz

kc)(bx

ka)λ(cy

ή

kλ)a(cy

k)(az

k)b(cx

b

with κ,λ in Q

Another form where, (n, f) = 1 with coprime integers as solutions of the equation, it is clear if we

replace the upper and 3 forms.

Proportionally controlled or if we want to only give positive or negative integer triples as

solutions.Another form where vfn ,, ,(n,f)=1 with coprime primes integers as solutions of

the equation,, it is clear if we replace f

n and vfk the upper and 3 forms.. Proportionally

controlled ε nf or nf if we want to only give positive or negative integer triples as

solutions.

ββ.. PPyytthhaaggoorreeaann ttrriipplleett sseeccoonnddaarryy iiss aallssoo aa lliisstt ooff tthhrreeee iinntteeggeerrss ((xx,, yy,, zz)) ssuucchh tthhaatt 222zyx (2)

. Such a trio called primary if the integers are non-negative and gcd(x,y,z)=1. For

example, (3,4,5) is a Pythagorean triple because primary 222543 . Although (30,40,50) and (-

3,4, -5) are also Pythagorean triples, but it is not primary. Using the same method we divide the

equation (2) with the y2, where 0y and we get

2

2

2

2

1y

z

y

x (3).

consider my

x and m

y

z where ( Qm , , Q and substituting in (3) should be

221)( mm but

2

11)(

2

22mmm και

2

12

m therefore choosing

2y , Zk implies that secondary Pythagorean triplet is

ky

kz

kx

)2(

)1(

)1(

2

2

( Ι)

For example if we take 2

1,4 k then implies that…

34

344))

2

1(1(

2x and 5

4

544))

2

1(1(

2z and 44)

2

12( y .

so the triple (3,4,5). Equation (I) has another form, is that we know from the old.

So we have the solution….

vnfy

vnfz

vnfx

)2(

)(

)(

22

22

(ΙΙ)

where vfn ,, ,(n,f)=1 with coprime integers as solutions of the equation and obtained by

replacing thef

n and

2vfk in (I) with nf and finally gives us only trebles as a positive

integer solutions. The primary Pythagorean triples with the following conditions:

1. 0 ≤ x ≤ y ≤ z,

2. gcd(x, y, z) = 1

3. 222zyx

until z <= 100 is::

(3,4,5) ,(5,12,13), (8,15,17), (7,24,25) , (20,21,29), (12,35,37), (9,40,41), (28,45,53), (11,60,61),

(33,56,65), (16,63,65),(48,55,73), (36,77,85), (13,84,85), (39,80,89), (65,72,97).

So we see that there are 16 primary Pythagorean triples with 0 ≤ x ≤ y ≤ z ≤ 100.

cc.. PPyytthhaaggoorreeaann ttrriipplleett tthhiirrdd ddeeggrreeee iiss kknnoowwnn aass aa lliisstt ooff tthhrreeee iinntteeggeerrss ((xx,, yy,, zz)) ssuucchh tthhaatt 333

zyx ((44)).. UUssiinngg tthhee ssaammee mmeetthhoodd wwee ddiivviiddee tthhee eeqquuaattiioonn ((44)) wwiitthh tthhee y3, wwhheerree 0y and

wwee ggeett 3

3

3

3

1y

z

y

x ((55))..

Consider my

x and m

y

z where ( Qm , , Q ) an by replacing in (5) will be

331)( mm but 0)1(331)(

32233 mmmm (6)

i) Solving (6) with respect to m,

6

)1(1293342

m , thus resulting the

minimum requirement in order to have integer solutions is the discriminant of (6) is

positive or zero.Thereforeς the minimum requirement in order to have integer

solutions is the discriminant of (6) is positive or zero. So Δ = 03124 whose

solutions are 340 .But Δ is positive in space ]4,0[

3 and because the full

discriminant must be positive integer then necessarily the current of 10 .

In that case, it should be that 1m and so yzx 0 . Pythagorean triplet

therefore will beη (x, y, z)= (0, y, y).

ii) We set the Below the root equal to 22

9 but then, however, should i.e..

4)3(22

.This implies that (λ,κ)=(1,1) which is like a unique solution. . As

previously shown in the case (1)

So as we see the third degree Pythagorean triplet is not susceptible generalized integer

solutions for all variables, especially since one necessarily equal to zero.

γγ.. PPyytthhaaggoorreeaann ttrriipplleett ffoouurrtthh ddeeggrreeee kknnooww tthhaatt iiss aa lliisstt ooff tthhrreeee iinntteeggeerrss (x, y, z) such that 444

zyx (7) . Using the same method we divide the equation (7) with y

4, where 0y and

we take 4

4

4

4

1y

z

y

x (8).

consider my

x and m

y

z where ( ,m , ) and substituting in (8) will be

441)( mm αλλά 0)1(4641)(

4322344 mmmmm (9) and

then

We see that the discriminant is positive for each . Obviously (9) holds when λ=1 thereby

resulting in m=1. But then yzx 0 the Pythagorean triplet therefore will be ….

(x, y, z)= (0, y, y).

CCoonnddiittiioonnss ffoorr ppoossiittiivvee ddiissccrriimmiinnaanntt aanndd ttoo ffiinndd tthhee uuppppeerr lliimmiitt ooff tthhee rroooottss ooff tthhee

ddiissccrriimmiinnaanntt..

i) . The λ is an integer in general..

Necessary and sufficient condition has integer solutions to the n-th degree Pythagorean triplet is

the discriminant of a solution to m, be positive, but at the same time the upper limit of the

variable λ , satisfies the condition 02 .

This should apply to prevent the removal of lambda, and to define the interval λ [0,2), which

necessarily means that apart from the price of λ=0 then

limited to the value λ=1 will be the ideal, but the only one that meets all the Conditions . To this

end, we examine two cases in detail:

11.. PPyytthhaaggoorreeaann ttrriipplleett ddeeggrreeee

,12

i) As we know, the discriminant of an equation is if we solve the system of equation and the

equation of the derivative. Thus we take as Δ the discriminant of generalized Equation odd

degree and with the specific, 01)(

mm με 12 ,and I will have the

replacement 01)(1212

mm (10). If the derivative the (10) and the equation to zero

we get 0)12()()12(22

mm (11).From (11) that comes 2

m and with the

replacement of (10) we find that the discriminant must be equal to

whifh

)2

(1)2

( 1212

)2

(1)2

( 12

)2

(21

which is

01))2

(21(lim12

1,

Which means that the discriminant Δ is positive for the price of 1,1 .

ii) To find the upper limit of the value of λ in the discriminant set the value of the Δ>=0 and find

the maximum because ))12/(1(

max

12)2

1(20)

2(21

which for tends to

the 2. So we obtain the value λ=1 as accepted solution. We will prove that there is no other

solution even any solution λ=-1 obtained from the equation 01)(

mm if we solve for

λ , and because we know that the relevant theorem with integer roots of a polynomial are the

divisors of the last term positive or negative ie. 1 . To this end we return to the original

equation 01)(

mm with ν=2ρ+1 and examine two cases if the solve for the variable

λ, so we have ...

Erect ...

1. λ=1

In this case, the original equation 01)(

mm becomes: :

01)1(1212

mm which has as unique solutions 1m .

i) m=1.

For this case fully verified 011)11(1212

fully verified .

ii) m=-1

For the case when 0)1(1)11(1212

verified when

0)1(1)2(12

which means that ρ=0 therefore necessarily ν=1.

2. λ=-1

In this case, the original equation 01)(

mm becomes::

01)1(1212

mm which has as unique solutions. 1m .

i) m=1

For the case 011)11(1212

can not be accepted because

22ρ+1

=0 which is impossible

ii) m=-1

For this case will force

02)1(1)0(12

which can not be so absurd. .

Reverse ....

To this end we return to the original equation

01)(

mm

with ν=2ρ+1 and examine two basic cases with respect to the m ...

1. m=1

In this case, the original equation 01)(

mm becomes:

011)1(1212

which has a unique solution 1 .

In this case the Pythagorean triplet is (x, y, z)= (0, y, y).

2. m=-1

Same as above, with the overall value the equation 01)(

mm becomes ι

2)1(0)1(1)1(121212

01 which means that ν=1.For ν>1

or ρ>0 the ratio leads to a contradiction, so there is only one solution for the λ so that ν=1 i.e.

λ=1.

In the given situation the trio is verified

(x, y, z)= (-2y, y, -y).

Therefore with 01)(

mm με ν=2ρ+1 is verified if applicable (m,λ)=(1,1) which

in this case is the Pythagorean triplet is (x, y, z)= (0, y, y) , 0y and for ν>1.

22..PPyytthhaaggoorreeaann ttrriiaadd ddeeggrreeee

,2

In a similar way the discriminant of an equation is if we solve the system of equation and the

equation of the derivative. So we take as Δ the discriminant equation generalized par grade and

specific 01)(

mm με 2 , and substituting then therefor i will have the

01)(22

mm (12). If the derivation of (12) and equation to zero should

0)(2)(21212

mm (13). From (13) we know that two equivalent where ...

analytically..

i)(m-λ)-m>=0 which implies -λ>=0 => λ<=0 hence we have the solutions m=-1 and λ=-1

ii)-(m-λ)-(-)m>=0 also which implies -λ<=0 => λ>=0 hence we have the solutions m=1 and

λ=1 excludes value λ=0.

And for the two cases have equivalent: :

0)1(1)11(0)1(1)11(2222

Which means that the discriminant is positive for the price 1 with no maximum .

To make what prices we receive for λ accepted as a solution, we prove that there is no other

solution of λ beyond of λ=-1 ή λ=+1 obtained from the equation 01)(

mm if we

solve for λ . We know that the relevant theorem integer roots of a polynomial are the divisors of

the last term positive or negative ie λ 1 . To this end we return to the original equation

01)(

mm with ν=2ρ and examine whether the two cases solve for the variable λ, so

we have ...

Erect ...

1. λ=1

In this case the initial equation 01)(

mm becomes::

01)1(22

mm which has as unique solutions 1m .

i) m=1

In this case 011)11(22

fully verified .

ii) m=-1

For this case 0)1(1)11(22

verified when

02)1(1)2(222

which means that ρ>=0 is impossible.

2. λ=-1

In this case the initial equation 01)(

mm becomes:

01)1(22

mm which has as unique solutions 1m .

i) m=1

For the case 011)11(22

can not be accepted because

22ρ

=0 , which is impossible

ii) m=-1

For this case will be

0)1(1)0(212

that applicable

Reverse ... for this let us return to the original equation with

01)(

mm with ν=2ρ and let us consider two basic cases with respect to the m ...

1. m=1

In this case the initial equation 01)(

mm becomes:

011)1(22

which has a unique solution. 1 . In this case the Pythagorean triplet is

(x, y, z)= (0, y, y).

2.m=-1

By replacing this general equation 01)(

mm becomes:

0)1(0)1(1)1(222

01 which means that ν>=2.

Certainly in the case only when ν = 2, the analysis will be given as before. In this case, with ν> 2

Pythagorean triplet is (x, y, z)= (0, y,-y).

To generalize the results should be mentioned theorem on irreducible fraction as a

polynomial solution, referred to evidence an explicit value of λ.

ii). The λ is a rational number in general.

THEOREM: If the polynomial equation 0...01

1

1

axaxaxa

n

with integer

coefficients has a solution as irreducible fraction κ/λ, ie (κ,λ)=1 , then κ is the divisor of 0

a and

the λ is divisor of

a .

Proof:

Having the κ/λ is a solution of the equation, we have

0)/(...)/()/(01

1

1

aaaa

0...

0

1

1

1

1

aaaa (1)

From (1),we have

0

1

1

2

1

1)...( aaaa

and because the κ divides the

1ο member and then I’ll divides and the second 2

ο member,i.e

0

/ a and after that,

1),( ,we ill have that 0

/ a .

From (1), and also we have

aaaa

)...(

1

0

2

2

1

1and because the λ

divides the 2ο second member and will divide and the 1

ο member ,i.e

a/ and after that

1),( , so we get that

a/ .

Therefore if we solve the original equation in the previous theorem 01)(

mm . we

will have…

i)ν=2ρ then has the general form

01)1(1)1(1

1

1

1

rr

r

кrr

r

кm

rmm

rm

Clearly if the root for λ is a rational the general form λ=p/q with (p,q)=1, I’ll we have

p=(+/-)1 and q=(+/-)1.Therefore λ=(+/-)1 and this because they are being negative and

positive as proved before..

ii)ν=2ρ+1 then has the general form

01)1(1)1(1

1

1

1

rr

r

кrr

r

кm

rmm

rm

Clearly if the root for λ is a rational the general form λ=p/q με (p,q)=1, I’ll be p=(+/-)1

and q=(+/-)1.Therefore λ=(+/-)1 and because one accepts only positive as shown then λ=1

and (p=+1,q=+1) as well (p=-1,q=-1). We conclude therefore for λ an integer number of

both cases. Therefore 01)(

mm με ν=2ρ,ν>2 is verified if applicable (m,λ)=(1,1)

which in this case is the Pythagorean triplet is (x, y, z)= (0, y, y) , 0y or if ν=2 and

Q ,1 the (x, y, z)= ((1-λ2)κ, 2λκ, ((1+λ

2)κ as originally shown.

Thereafter if applicable (m,λ)=(-1,1) με ν>2, the triad will be verifiable by (x, y, z)= (0, y, -

y). But accept because x>=0, y>0, z>=0 then finally accept only positive, so the couple

(m,λ)=(1,1) and more specifically for ν >2.

CCoonncclluussiioonn ......

II..AAppooddeeiikknnyyeettaaii ffuullllyy tthhaatt tthhee FFeerrmmaatt eeqquuaattiioonn aanndd eexxhhiibbiittoorrss 12 wwiitthh νν>>11 tthhee

eeqquuaattiioonn FFeerrmmaatt nnoo hhaass ssoolluuttiioonn ooff ttrriiaadd,, oouuttssiiddee 33 ggeenneerraall ccaasseess--ttiioonnss::

1.x=0,(x, y, z)= (0, y, y)

2.y=0, (x, y, z)= (χ, 0, χ)

3.z=0, i.(x, y, z)= (x, -x, 0)

ii. (x, y, z)= (-y, y, 0)

II. For the case where 2 και ν>2 the equation has no solution with relatively with λ,m

namely triad (x, y, z)= (0, (+/-)y,(+/-)y) with χ=0, or the symmetric (x, y, z)= ((+/-)x,0,(+/-)x)

with y=0, for (λ,m)=(1,1). Especially when ν=2 the only triad that is verifiable and showed

accordance with the foregoing is the (x, y, z)= ((1-λ2)κ2, 2λκ2, (1+λ

2)κ2) or another format

first on each other, the (x, y, z)= ((f2-n

2)κ2, 2fnκ2, (f

2+n

2)κ2).

4. The obvious solution is the (x, y, z)= (0, 0, 0).

Valid for positive integer solutions that:

Zzyx ,, ,f,n,

21

, kk , Qm , .

References

[1]. The Proof Of Fermat’s Last Theorem, Nigel Boston 2003

[2].Solve Equations, Mantzakouras Nikos 2014