A Error in (Multi)linear Regression

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Hi,
I keep reading varying accounts on conditions needed to " justify" the use of ( multi) linear regression to model data.

Specifically, I have seen several authors require errors to be normal, i.i.d , whilr others only require the errors be i.i.d with mean 0. Just where is the assumption of normality used to justify the use of linear models? I know of Gauss Mark of, but this seems too strong. I've heard that it is used to find the distribution of the coefficients and determine reliable confidence intervals for the coefficients? If so, do you suggest a source? If not, can you explain?
 
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I guess the question is what does "justify" mean. If the errors are zero mean iid normal and equally distributed, then I think the multi linear model is an unbiased estimator of the actual values. If the errors are otherwise distributed then a different model (i.e. different coefficients) might be a better choice.

But in practice, a multi linear model works pretty well in lots of other situations, and it's often hard to know that the errors actually look like.
 
Thank you. Just curious, as an aside, do you know of situations that cannot be modeled (log)linearly; with linearity meaning linearity in the coefficients?
 
You just mean give a scenario where a non linear model is better? Sure, suppose you drop an object, and write down the time and the height at those times as it falls. The height measurement has some noise to it. Then the right model for the height is going to be something like ##h(t)=-\frac{g}{2}t^2+h_0## if it starts at a height of ##h_0##.

I feel like there's a good chance I did not understand the question.
 
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This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
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