Uc 1: Fill & Download for Free

There are several timelines so let’s start with the most prolific, the Universal Century, some series overlap in terms of times and / or rehashes, the other timelines can be watched in any order as they don’t relate.Gundam Origin (Retelling of Mobile Suit Gundam, but starting before the ‘one year war’) (Still episodes to be aired)Mobile Suit Gundam08th MS teamGundam Thunderbolt (Still 2 more EP’s to be aired at time of writing, set during the one year war)Zeta GundamZZ Gundam (Double Zeta)0080 War in the pocket0083 Stardust memoriesMS Igloo (Various OVA’s, various times before 0083)Char’s Counterattack (0093)Gundam UnicornGundam F91Victory GundamTurn A GundamThen there is the CE (Celestial Era, Future UC):Gundam SEEDGundam SEED DestinyGundam SEED StargazerThen RG (Reguild Century, future UC)Reconguista in GThen AW (After War, alternative UC)Gundam XThen AC (After Colony)Gundam WingGundam Win Endless WaltzThen AD (Anno Domini)Gundam 00 seasons 1 + 2Gundam 00 Awakening of the TrailblazerThen FC (Future Century)G-GundamThen PD (post Disaster)Iron Blooded OrphansNoteable but not entirely Gundam Universe or Meta-Gundam:Gundam Build FightersGundam Build Fighters TriGundam Evolve (Various short films from different timelines)There’s probably some more missing off this list, but it’s enough to get you started on a very long binge watch!

First, set [math]x + y = a, x + z = b, y + z = c[/math].Then whenever [math]x,y,z[/math] are rational, so are [math]a,b,c[/math]. Also, we have[math]x = \displaystyle\frac{a+b-c}{2}, y = \displaystyle\frac{a+c-b}{2}, z = \displaystyle\frac{b+c-a}{2}[/math]That is, picking [math]a,b,c[/math] as rational ensures that [math]x,y,z[/math] are so as well. Thus it is enough to find rational solutions for[math]a^2 + b^2 + c^2 = 2[/math]There is a standard trick to find solutions to such problems. We start with finding one solution. In this case, [math](a_0,b_0,c_0) = (1,1,0)[/math] can be found easily.Think of [math](a,b,c)[/math] as the three-dimensional space. [math]a^2 + b^2 + c^2 = 2[/math] is then the equation of the sphere with radius [math]\sqrt{2}[/math].The first solution is marked as [math]\mathbf{r}_0[/math].Now, consider any other rational point on the sphere [math]\mathbf{r}=(a',b',c')[/math].Unless its [math]c[/math]-coordinate is zero, it has coordinates in the general form[math](1+uc',1+vc',c') \tag{1}[/math]where [math]u[/math] and [math]v[/math] are the slopes of the line joining [math]\mathbf{r}_0[/math] and [math]\mathbf{r}[/math].Now, it is easy to see that if [math]\mathbf{r}[/math] is a rational point, then [math]u = \frac{a'-1}{c'}[/math] and [math]v = \frac{b'-1}{c'}[/math] are rational as well.Similarly, suppose that [math]u[/math] and [math]v[/math] are arbitrary rational numbers. The line with slopes [math]u[/math] and [math]v[/math] will have coordinates [math](1+uc',1+vc',c')[/math] and will intersect the sphere when the condition [math](1+uc')^2 + (1+vc')^2 + c'^2 = 2[/math] holds. Solving this gives[math]c' = -\displaystyle\frac{2(u+v)}{u^2+v^2+1} \tag{2}[/math]That is, the coordinates are guaranteed to be rational as well. This gives us a bijection between arbitrary pairs of rational numbers [math](u,v)[/math] and rational points on the sphere [math]r[/math]. Plugging the value of [math]c'[/math] from [math](2)[/math] in [math]([/math][math]1[/math][math])[/math], we have the general solution[math]a = \displaystyle\frac{1-u^2-2uv+v^2}{u^2+v^2+1}[/math][math]b = \displaystyle\frac{1+u^2-2uv-v^2}{u^2+v^2+1}[/math][math]c = -\displaystyle\frac{2(u+v)}{u^2+v^2+1}[/math]From this, the values of [math](x,y,z)[/math] can be found as[math]x = \displaystyle\frac{1+u+v-2uv}{u^2+v^2+1}[/math][math]y = \displaystyle\frac{v^2-u^2-u-v}{u^2+v^2+1}[/math][math]z = \displaystyle\frac{u^2-v^2-u-v}{u^2+v^2+1}[/math]This finds almost all solutions, except the ones with [math]c=0[/math]. But that is equivalent to solving the reduced dimensional problem of finding the rational solutions of [math]a^2+b^2 = 2[/math]. The same procedure as we followed applies, with the sphere getting modified to a circle. Solving the very similar equation [math]a^2+b^2=1[/math] is one way to generate all pythagorean triples.[1] Try it out!Footnotes[1] Parametrization of Pythagorean triples

UC 1 - 3 costed approximately $20 million each to make which is considerably a small figure for such fine games .Naughty Dog is known for their frugal use of resources and are one of the proponents of ‘less is more’ . No ready figures have been revealed by the developers yet, But the rumors state that it is considerably lesser than $50 million.