__The Proof of FLT and Beal conjecture in 5 steps__
M.Meyyappan#

**#**(Professor of Physics (Retd), Alagappa Government Arts College, Karaikudi-630002, India}

*Abstract:*
Both the Fermat’s
Last Theorem and Beal conjecture can be proved in five steps.

*Key words***:**

FLT, Beal Conjecture, multi-power relations, number
theory

**Introduction**

**In literature quite a large number of references are available for Fermat’s Last Theorem and Beal conjecture [1]{2}{3}. The Pythagoras theorem, FLT and Beal conjecture are different form of one and the same multi-power relation under different condition. The most general form of three member multi-power relation is x**

^{p }+ y

^{q }= z

^{r }, where both the base numbers (x,y,z) and the exponents (p,q,r) are all supposed to be whole integers. Under different conditions, it gives Pythagorean relation (p = q = r = 2), Fermat last theorem (p = q = r = n and n > 2) and Beal conjecture (p,q,r are not equal, at least one is different and p = q = r ≠ 2).

When x,y,z,p,q and r
are positive integers with p,q,r are greater than 2, then x,y,z will invariably
have a common prime factor, when the exponent of one or two base numbers is 2, the common prime
factor is not to be a necessary condition and when the exponents of all the
base numbers are equal, the common prime factor present can be cancelled out
and reduces ultimately to an irreducible form of the relation. In this paper a
simple proof for FLT and Beal conjecture is derived from the realistic
Pythagoras theorem.

**1.FLT in Five Steps**

Step.1,Consider any
known Pythagorean triples (a,b,c) satisfying a

^{2 }+ b^{2 }= c^{2 , }where a,b and c are all whole integers. Dividing by a^{2 }throughout gives reduced Pythagorean relation as 1 + (b/a)^{2 }= (c/a)^{2}
Step,2 Let us suppose
an unknown three member like power relation
x

^{p}+ y^{p }= z^{p, }where x,y and z may or may not be whole integers,
Step 3. Multiplying
throughout the reduced Pythagorean relation by x

^{p }gives x^{p }+ x^{p }(b/a)^{2 }= x^{p }(c/a)^{2}
Step 4. Comparing term
by term with the unknown relation y

^{p }= x^{p }(b/a)^{2}and x^{p }(b/a)^{2 }+ k = (xα)^{p}^{ }and z^{p }= x^{p }(c/a)^{2}and x^{p }(c/a)^{2}+ k^{ }= (xβ)^{p }where α and β are assumed to be whole integers and k will be zero when (b/a)^{2 }= α^{p }and (c/a)^{2 }= β^{p}if not k may have some positive or negative value
Step 5. Substituting
the derived values for (b/a)

^{2 }and (c/a)^{2 }in the reduced Pythagorean relation 1 + α^{p }= β^{p }. No integer or fractional values of α and β will satisfy this relation for all values of p except p = 2.**II. Beal conjecture in Five steps**

**2**

Step.1 Consider any
known Pythagorean triples (a,b,c) satisfying a

^{2 }+ b^{2 }= c^{2 , }where a,b and c are all whole integers. Dividing by a^{2 }throughout gives reduced Pythagorean relation as 1 + (b/a)^{2 }= (c/a)^{2 }
Step 2. Let us suppose
an unknown three member unlike power relation
x

^{p}+ y^{q }= z^{r }where x,y,z,p,q and r are supposed to be whole integers.
Step 3. Multiplying
throughout the reduced Pythagorean relation by x

^{p }gives x^{p }+ x^{p }(b/a)^{2 }= x^{p }(c/a)^{2}
Step 4.Comparing term
by term with the unknown relation, y

^{q }= x^{p }(b/a)^{2 }= (xα)^{p }and z^{r }= x^{p }(c/a)^{2}= x^{p+k }where α , β and k are all assumed to be whole integers.
Step 5. Substituting
the derived values for (b/a)

^{2 }and (c/a)^{2 }in the reduced Pythagorean relation 1 + α^{p }= x^{k. }. Few possible solutions and the corresponding multi-power relations are 1 + 2^{3 }= 3^{2 }gives 3^{3 }+ 6^{3 }= 3^{5 , }1+ 7 = 2^{3 }gives 7^{3 }+ 7^{4 }= 14^{3 }and few general solutions are 1 + (2^{n }– 1} = 2^{n }gives (2^{n }- 1)^{n }+ ( 2^{n }-1)^{n+1 }= [2(2^{n }-1)]^{n }and 1 + n^{m }= (n^{m }+ 1) gives ( n^{m }+ 1)^{m }+ [n(n^{m }+1)]^{m }= (n^{m }+ 1)^{m+1}.**III. Conclusion:**

The proof of FLT and
Beal conjecture in five steps shows that three member equal power relation is
not possible with all the base numbers are integers for all exponents greater
than 2 since 1 + α

^{n }≠ β^{n}.However such relations are possible when exponents of the base numbers are allowed to be different since 1 + α^{n }= β^{m}.When it is multiplied with a common multiplier (β^{n}) which becomes the common prime factor in Beal conjecture we get a multi-power relation β^{n }+ (αβ)^{n }= β^{m+n, }while the common factor α^{m }gives α^{m }+ α^{ m+n }= (αβ)^{m}.**References**

**[1] M.Meyyappan http://www.ijmttjournal.org/2017/volume-45number-1/IJMTT-V45P503**

**[2]Gerd Falting,”The Proof of Fermat’s Last Theorem by R.Taylor and A,Wills” Notices AMS, 1995,42(7)743—746.**

**[3]Simon Singh, “Fermat’s Last Theorem” Fourth Estate Ltd., 1997**