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edit approved Jul 29, 2015 at 18:15

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Here is how Fermat probably did it (it is how I did it - not all of the steps were needed but I have to believe this was close to Fermat's thought process).

Any prime of the form 8n+1$8n+1$ or 8n+3$8n+3$ can be written in the form a^2 +2b^2$a^2 +2b^2$. This is proved with descent techniques once realizes that -2$-2$ and 1$1$ are squares mod 8n+1$8n+1$ or 8n+3$8n+3$ and hence setting a^2=-2$a^2=-2$ and b^2 = 1$b^2 = 1$ gets the result of 0$0$ (mod 8n+1$8n+1$ or 8n+3$8n+3$) for a^2+2b^2$a^2+2b^2$, which means our prime divides the result. Any prime of the form 8n+5$8n+5$ or 8n+7$8n+7$ cannot be.

Point two is that combinations of squares with common shapes when multiplied by each other retain their shape. Let x = a^2 + Sb^2$x = a^2 + Sb^2$, and y = c^2 + Sd^2$y = c^2 + Sd^2$. xy = (ac+Sbd)^2 + S(ad-bc)^2 = (ac-Sbd)^2 + S(ad+bc)^2$xy = (ac+Sbd)^2 + S(ad-bc)^2 = (ac-Sbd)^2 + S(ad+bc)^2$

Point three is that if y$y$ is even y^2 + 2$y^2 + 2$ is even as is x^3$x^3$. Dividing both sides by 2$2$ would make the left hand side odd and right hand side even so both y$y$ and x$x$ are odd.

Point four is that if a non-prime is of the form a^2 + 2b^2$a^2 + 2b^2$ then all its prime factors must be of the form 8n+1$8n+1$ or 8n+3$8n+3$, or the factor must be a square.

Point five is to observe that y^2 + 2$y^2 + 2$ is of the form a^2 + 2b^2$a^2 + 2b^2$ with a=y$a=y$ and b=1$b=1$. Combining this with four and one means there are no squares of the form 8n+5$8n+5$ or 8n+7$8n+7$ since b$b$ would be equal to that square, not 1$1$.

So now we expand upon point three to make the proof. x $x$ is of the form a^2 + 2b^2$a^2 + 2b^2$. x^3 $x^3$ can be written as (a^3-3Sab^2)^2 + S(3a^2b-Sb^3)^2$(a^3-3Sab^2)^2 + S(3a^2b-Sb^3)^2$. Letting S=2$S=2$ we see that the expression (3a^2b-2b^3)^2$(3a^2b-2b^3)^2$ must be equal to 1$1$. Hence b^2 * (3a^2-2b^2)^2 =1$b^2 \cdot (3a^2-2b^2)^2 =1$. Using positive integers we see b=a=1 $b=a=1$ is the only solution. Hence x =1^2 + 2*1^2 = 3$x =1^2 + 2*1^2 = 3$ is the only possibility and 5^2 + 2 =3^3$5^2 + 2 =3^3$ is the only solution

Here is how Fermat probably did it (it is how I did it - not all of the steps were needed but I have to believe this was close to Fermat's thought process).

Any prime of the form 8n+1 or 8n+3 can be written in the form a^2 +2b^2. This is proved with descent techniques once realizes that -2 and 1 are squares mod 8n+1 or 8n+3 and hence setting a^2=-2 and b^2 = 1 gets the result of 0 (mod 8n+1 or 8n+3) for a^2+2b^2, which means our prime divides the result. Any prime of the form 8n+5 or 8n+7 cannot be.

Point two is that combinations of squares with common shapes when multiplied by each other retain their shape. Let x = a^2 + Sb^2, and y = c^2 + Sd^2. xy = (ac+Sbd)^2 + S(ad-bc)^2 = (ac-Sbd)^2 + S(ad+bc)^2

Point three is that if y is even y^2 + 2 is even as is x^3. Dividing both sides by 2 would make the left hand side odd and right hand side even so both y and x are odd.

Point four is that if a non-prime is of the form a^2 + 2b^2 then all its prime factors must be of the form 8n+1 or 8n+3, or the factor must be a square.

Point five is to observe that y^2 + 2 is of the form a^2 + 2b^2 with a=y and b=1. Combining this with four and one means there are no squares of the form 8n+5 or 8n+7 since b would be equal to that square, not 1.

So now we expand upon point three to make the proof. x is of the form a^2 + 2b^2. x^3 can be written as (a^3-3Sab^2)^2 + S(3a^2b-Sb^3)^2. Letting S=2 we see that the expression (3a^2b-2b^3)^2 must be equal to 1. Hence b^2 * (3a^2-2b^2)^2 =1. Using positive integers we see b=a=1 is the only solution. Hence x =1^2 + 2*1^2 = 3 is the only possibility and 5^2 + 2 =3^3 is the only solution

Here is how Fermat probably did it (it is how I did it - not all of the steps were needed but I have to believe this was close to Fermat's thought process).

Any prime of the form $8n+1$ or $8n+3$ can be written in the form $a^2 +2b^2$. This is proved with descent techniques once realizes that $-2$ and $1$ are squares mod $8n+1$ or $8n+3$ and hence setting $a^2=-2$ and $b^2 = 1$ gets the result of $0$ (mod $8n+1$ or $8n+3$) for $a^2+2b^2$, which means our prime divides the result. Any prime of the form $8n+5$ or $8n+7$ cannot be.

Point two is that combinations of squares with common shapes when multiplied by each other retain their shape. Let $x = a^2 + Sb^2$, and $y = c^2 + Sd^2$. $xy = (ac+Sbd)^2 + S(ad-bc)^2 = (ac-Sbd)^2 + S(ad+bc)^2$

Point three is that if $y$ is even $y^2 + 2$ is even as is $x^3$. Dividing both sides by $2$ would make the left hand side odd and right hand side even so both $y$ and $x$ are odd.

Point four is that if a non-prime is of the form $a^2 + 2b^2$ then all its prime factors must be of the form $8n+1$ or $8n+3$, or the factor must be a square.

Point five is to observe that $y^2 + 2$ is of the form $a^2 + 2b^2$ with $a=y$ and $b=1$. Combining this with four and one means there are no squares of the form $8n+5$ or $8n+7$ since $b$ would be equal to that square, not $1$.

So now we expand upon point three to make the proof. $x$ is of the form $a^2 + 2b^2$. $x^3$ can be written as $(a^3-3Sab^2)^2 + S(3a^2b-Sb^3)^2$. Letting $S=2$ we see that the expression $(3a^2b-2b^3)^2$ must be equal to $1$. Hence $b^2 \cdot (3a^2-2b^2)^2 =1$. Using positive integers we see $b=a=1$ is the only solution. Hence $x =1^2 + 2*1^2 = 3$ is the only possibility and $5^2 + 2 =3^3$ is the only solution

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created Jul 29, 2015 at 17:41

Bob

Here is how Fermat probably did it (it is how I did it - not all of the steps were needed but I have to believe this was close to Fermat's thought process).

Point three is that if y is even y^2 + 2 is even as is x^3. Dividing both sides by 2 would make the left hand side odd and right hand side even so both y and x are odd.

Point four is that if a non-prime is of the form a^2 + 2b^2 then all its prime factors must be of the form 8n+1 or 8n+3, or the factor must be a square.