3 Fixed a typo.

I have an English translation of the letter; you can find my email address on my homepage. Here is the relevant part:

"By the way, I take it to be not useless to note down also such propositions that are very probable, even if a real proof is lacking; for even if afterwards they were found to be erroneous, they could all the same give occasion for the discovery of a new truth. Thus Fermat's idea that all the numbers $2^{2^{n-1}}+1$ yield a series of prime numbers cannot hold up, as you already demonstrated, Sir; but it should still be remarkable if this series were composed only of numbers that could be split into two squares in a unique way. I should like to risk another conjecture of that kind: any number composed from two primes is the sum of as many squares of prime numbers (including $1$) as one wishes, right down to the sum that consists just of ones.

After reading this through again, I see that the conjecture can be proved quite rigorously for the case $n+1$ if it holds for the case $n$ and $n+1$ can be split into two prime numbers. The proof is very easy; and at least it appears to be true that every number greater than $2$ is the sum of three prime numbers."

The translation was done by Martin Mattmüller.

2 I included the translation of parts of the letter.

I have an English translation of the letter; you can find my email address on my homepage. Here is the relevant part:

"By the way, I can take it to be not useless to note down also put such propositionsthat are very probable, even if a real proof is lacking; for even ifafterwards they were found to be erroneous, they could all the relevant passage here same giveoccasion for the discovery of a new truth. Thus Fermat's idea that all thenumbers $2^{2^{n-1}}+1$ yield a series of prime numbers cannot hold up, asyou already demonstrated, Sir; but it should still be remarkable if desiredthisseries were composed only of numbers that could be split into two squares ina unique way. I should like to risk another conjecture of that kind: any number composed from two primes is the sum of as many squares of prime numbers (including $1$) as one wishes, right down to the sum that consists just of ones.

After reading this through again, I see that the conjecture can be proved quite rigorously for the case $n+1$ if it holds for the case $n$ and $n+1$ can be split into two prime numbers. The proof is very easy;and at least it appears to be true that every number greater than $2$ is the sum of three prime numbers."

The translation was done by Martin Mattmüller.

1

I have an English translation; you can find my email address on my homepage. I can also put the relevant passage here if desired.