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I have constructed the number $7131372917538397234773191167617941438959$, which is prime as shown here, using all primes under $100$ which contains two digits by randomly ordering them as $31,37,29,\ldots, 59$ except $13$. I have got that number $7131372917538397234773191167617941438959$ satisfies the following properties:

1.- The sum of its digits is also prime: it is equal to $193$.

3.- The number is of the form $6n+1$

4.- This number can't be written as a sum of $3$ squares.

Now my question here is:

Could be this : $7131372917538397234773191167617941438959$ written as $x^{2}+y^{2}$ with $x, y$ integers?

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    The only information here is that it's prime and the last two digits. – 2017-02-08
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    Related: http://mathoverflow.net/questions/261694 – 2017-02-08
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    yes, i know and it is my question there is not well received but here it's not the same question then why this downvote ? – 2017-02-08
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    The usual criterion for three squares includes the possibility of one of the squares being zero. So if M is not a sum of 3 squares, it isn't a sum of 2 squares either. [I'm assuming you checked usual criterion for sum of 3 squares...] – 2017-02-08
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    I suppose people consider this question to not be interesting for precisely the same reasons the other one wasn't considered to be. The only difference is that in this case the answer is easy to find. – 2017-02-08
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    Try 71313729175383972347731911676179414389597131372917538397234773191167617941438961. Much easier. – 2017-02-08

1 Answers 1

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No, it cannot. Note that for all $x \in \mathbb{Z}$, $$x^2 \equiv 0, 1 \pmod {4}$$ Thus a sum of two squares can only be $0,1,2$ modulo $4$.

However, note that $$7131372917538397234773191167617941438959 \equiv 59 \equiv 3 \pmod {4}$$ Thus, it is not a sum of two squares. We are done!

For primes that are the sum of two squares, see Fermat's Theorem on the Sum of two Squares. For a prime $p$ to be of the form $$p=x^2+y^2$$ Then $p \equiv 1 \pmod {4}$ or $p=2$.

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    One doesn't even need to invoke primality nor Fermat's theorem - no number which is $3\pmod 4$ is a sum of two squares, since squares mod $4$ are $0,1$. – 2017-02-08
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    @Wojowu OK. But wouldn't this provide more information to the OP? – 2017-02-08
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    I agree this provides more info. I am just saying that we don't need it. – 2017-02-08
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    as this could be written as a sum of 2 squares :171313729175383972347731911676179414389593, the same number but i add 3 in the right and 1 in the left of the titled number – 2017-02-08
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    @Wojowu, i think the interesting is to ask if the titled number is a Mersann prime !!! – 2017-02-08
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    @zeraouliarafik But think about it this way - there are billions of billions of primes smaller than the exponent you have taken. There is little reason to think it's easier to answer your question for _this particular exponent_ than it would be for any other one. Your number is just fairly arbitrary. Let me point out to you that there are hundreds of other permutations of primes you could've chosen; why this particular one? – 2017-02-08
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    @zeraouliarafik And the number in your above comment is $1 \pmod {4}$, so it is the sum of two squares if it is prime. – 2017-02-08
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    @zeraouliarafik Also, note that that is already a MO question. – 2017-02-08
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    yes it is my question in MO but it's not received there , but here is other easy question after your answer – 2017-02-08
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    @zeraouliarafik Well, ask it as a seperate question. It is answered in MO, as well. – 2017-02-08