Permalink Submitted by Anonymous on August 24, 2011

Fermat was keen to prove his conjecture (FLT) for exponents 3, 4 & 5 because once this is achieved then it automatically extends to all integer exponents. My guess at the "marvelous proof" claimed by Fermat is as follows:
Let us imagine solid unit cubes (of side unity) to represent the number '1'. Then 'x' of these would represent the number 'x' and let us imagine these are placed in a linear array. We can then place 'y' of these unit cubes to represent the number 'y'.
We can similarly contruct 'x^3' (cube of 'x') using the unit cubes by extending the linear array to a square array and finally 'x' layers of the square array to form the cube - 'x^3'. The cube of 'y' can be similarly contructed and placed alongside the cube of 'x'.

FLT asserts that the sum of the cubes of 'x' and 'y' cannot be equal to another cube, say of 'z'.

Now, 'x^4' - the exponent 4 of 'x' - can be constructed by a linear array of the cubes of 'x' - taking 'x' numbers of the cubes of 'x'. Similarly we take 'y' numbers of the cubes of 'y' and place these two linear arrays of 'x^3' and 'y^3' alongside.
Then the exponent 5 for 'x' and 'y' would be represented by square arrays of the cubes of 'x' and 'y'.

FLT states that the sum of 'x^4' and 'y^4' cannot be equal to any 'z^4'. Similarly for exponent 5. Also, there exists proof for exponents 3, 4 & 5.

Finally, the exponent 6 for 'x' and 'y' will turn the square arrays of cubes into "super-cubes"!! And exponent 7 would be a repeat of exponent 4 as above; exponent 8 would be a repeat of 5 and exponent 9 will again form "super-super-cubes", that is, essentially exactly like the first cube (x^3)!

Therefore if FLT is true for exponents 3, 4 & 5 then it is true for exponents 6, 7 & 8 and again for 9, 10 & 11 and so on for all the integers!

Please tell me if this holds water or is there a flaw in my reasoning?
Jimmy A. M. Kazmi, Hyderabad, India

## Fermat's Last Theorem

Fermat was keen to prove his conjecture (FLT) for exponents 3, 4 & 5 because once this is achieved then it automatically extends to all integer exponents. My guess at the "marvelous proof" claimed by Fermat is as follows:

Let us imagine solid unit cubes (of side unity) to represent the number '1'. Then 'x' of these would represent the number 'x' and let us imagine these are placed in a linear array. We can then place 'y' of these unit cubes to represent the number 'y'.

We can similarly contruct 'x^3' (cube of 'x') using the unit cubes by extending the linear array to a square array and finally 'x' layers of the square array to form the cube - 'x^3'. The cube of 'y' can be similarly contructed and placed alongside the cube of 'x'.

FLT asserts that the sum of the cubes of 'x' and 'y' cannot be equal to another cube, say of 'z'.

Now, 'x^4' - the exponent 4 of 'x' - can be constructed by a linear array of the cubes of 'x' - taking 'x' numbers of the cubes of 'x'. Similarly we take 'y' numbers of the cubes of 'y' and place these two linear arrays of 'x^3' and 'y^3' alongside.

Then the exponent 5 for 'x' and 'y' would be represented by square arrays of the cubes of 'x' and 'y'.

FLT states that the sum of 'x^4' and 'y^4' cannot be equal to any 'z^4'. Similarly for exponent 5. Also, there exists proof for exponents 3, 4 & 5.

Finally, the exponent 6 for 'x' and 'y' will turn the square arrays of cubes into "super-cubes"!! And exponent 7 would be a repeat of exponent 4 as above; exponent 8 would be a repeat of 5 and exponent 9 will again form "super-super-cubes", that is, essentially exactly like the first cube (x^3)!

Therefore if FLT is true for exponents 3, 4 & 5 then it is true for exponents 6, 7 & 8 and again for 9, 10 & 11 and so on for all the integers!

Please tell me if this holds water or is there a flaw in my reasoning?

Jimmy A. M. Kazmi, Hyderabad, India