## Tuesday, May 30, 2017

### creative thoughts

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 {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 xp  + y = zr , 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 a2  + b2  = c2 ,  where a,b and c are all whole integers. Dividing by athroughout gives reduced Pythagorean relation as 1 + (b/a)= (c/a)2
Step,2 Let us suppose an unknown three member like power relation  xp + y= zp,  where x,y and z may or may not be whole integers,
Step 3. Multiplying throughout the reduced Pythagorean relation by xp  gives  xp  + xp (b/a)2  = xp (c/a)2
Step 4. Comparing term by term with the unknown relation  yp  = xp (b/a)2 and xp (b/a)2  + k =  (xα)p   and zp  = xp (c/a)2 and xp (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  =  αp  and (c/a)2  = βp if not k  may have some positive or negative value
Step 5. Substituting the derived values for (b/a)and (c/a)2 in the reduced Pythagorean relation 1 + αp =    βp . 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 a2  + b2  = c2 ,  where a,b and c are all whole integers. Dividing by athroughout gives reduced Pythagorean relation  as 1 + (b/a)= (c/a)
Step 2. Let us suppose an unknown three member unlike power relation  xp + yq  = zwhere x,y,z,p,q and r are supposed to be whole integers.
Step 3. Multiplying throughout the reduced Pythagorean relation by xgives  xp  + xp (b/a)2  = xp (c/a)2
Step 4.Comparing term by term with the unknown relation, yq   = xp (b/a)2 = (xα)p  and zr  = xp (c/a)2 = xp+k  where α , β and k are all assumed to be whole integers.
Step 5. Substituting the derived values for (b/a)and (c/a)2 in the reduced Pythagorean relation 1 + αp  = xk. . Few possible solutions and the corresponding multi-power relations are 1 + 2 = 32 gives  33  + 63  = 35 ,  1+ 7 = 2gives  73  + 74 = 143  and few general solutions are  1 + (2n – 1} = 2n gives  (2n  - 1)n   + ( 2n -1)n+1  = [2(2n -1)]n  and 1 + nm  = (nm + 1)  gives ( nm  + 1)+ [n(nm +1)]m  = (n+ 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 + αn  ≠ βn .However such relations are possible when exponents of the base numbers are allowed to be different since 1 + αn  = βm.When it is multiplied with a common multiplier (βn) which becomes the common prime factor in Beal conjecture we get a multi-power relation βn  + (αβ)n  = βm+n, while the common factor αm  gives αm  + α m+n  = (αβ)m .
References
 M.Meyyappan http://www.ijmttjournal.org/2017/volume-45number-1/IJMTT-V45P503
Gerd Falting,”The Proof of Fermat’s Last Theorem by R.Taylor and A,Wills” Notices AMS,          1995,42(7)743—746.

Simon Singh, “Fermat’s Last Theorem” Fourth Estate Ltd., 1997