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There is a unit u and primes $\pi_{j}=a_{j}+b_{j}i$ in $\mathbb{Z}[i]$ with $a_{j}>0, b_{j}>0$ and $7+i = u\pi_{1}\dots\pi_{k}$

$\mathbb{Z}[i]$ has four units: $i,-i,1,-1$. The product of the primes is also in $\mathbb{Z}[i]$, so : $7+i=u\pi… \pi_{k} = i(a+bi) = -b+ai$. This can't be the case since then it would follow that : $a=1, b=-7$ but since $b_{j}> 0$ i therefore can't be that unit.
For -i : we get -i(a+bi)= -ai+b . With this it follows that $a=-1, b= 7 $, therefore since $a_{j}>0$ this also can not be true.
For 1: 1(a+bi)= a+bi , a=7, b=1. This could be true, so now it is to show that 7+i is prime in $\mathbb{Z}[i]$. We can see directly that 7+i is irreducible over $\mathbb{Z}[i]$ and that means it must also be prime since $\mathbb{Z}[i]$ is a UFD. Therefore 1 is that unit.

I believe this can't be right, because I didn't use the indices at all. Does anybody see the right way? Please do tell.

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    Thanks to all of you. Result is: $u=-i , (1+i)(2+i)(2+i)$2011-11-16

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Your claim that $u=i$ is impossible does not follow.

You know that each $b_i$ is positive; but that does not tell you that the final product $a+bi$ must have $b$ positive.

For example, you can have $(a_1+b_1i)(a_2+b_2i)$ with $b_1,b_2\gt 0$, but with product $(a_1a_2-b_1b_2) + (a_1b_2+b_1a_2)i$ having imaginary part negative. Just take $a_1=0$, $b_1=2$, $a_2=-2$, $b_2=1$. Then you have $(0+2i)(-2+i)$, and the product is $-2-4i$, with negative real and negative imaginary parts. The other conclusions are likewise incorrect.

Finally, your claim that $7+i$ is irreducible is also incorrect. Note that $N(7+i) = 50 = 2\times 5^2$; if it is reducible, you could look for an element with norm $5$ and one with norm $10$, which quickly leads to $7+i = (2+i)(3-i)$. Since neither factor is a unit (their norms are indeed $5$ and $10$), then this shows $7+i$ is indeed reducible. (You can also look for elements with norms $2$ and $25$).

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    Arturo Magidin, Thanks.2011-11-16