Like in the title, I have to check that Goldbach Conjecture holds for every even number between 900 and 1000.
It's worth to note that I'm not supposed to use computer, so writing e.g. Python code will not do the trick.
Only idea that's coming to my mind is to check every number manually, but it looks very inefficient and like a kind of non-sens.
Is there any better way to do that? (I doubt it to be honest, but hopes I'm wrong)
On
I cheated, but you could do this by hand easily enough. Make up $2$ sieve tables, one from $0$ to $79$ and the other from $870$ to $1009$. Going through the first table staring from $3$, replace all of its odd multiples in both tables with the current prime. Then the next odd number in the first table which hasn't been replaced is prime, so replace all of its odd multiples and so on until a pass through the first table has been completed.
At this point you know all the primes in both tables. Then in the second (and optionally also the first) table subtract all primes in the first table from each even number and when the difference is a prime visible in the table, replace the current number with the first prime in the first table for which this was the case.
When you are done you should have $4$ classes of numbers: odd primes, smallest divisors of odd numbers, smallest prime which is the difference between the current number and any odd prime in the tables, and the rest. The !@#\$%^&* number here turns out to be $992$. $$\begin{array}{cccccccccc} 0&1&2&\color{green}{3}&4&\color{green}{5}&\color{blue}{3}&\color{green}{7}&\color{blue}{3}&\color{red}{3}\\ \color{blue}{3}&\color{green}{11}&\color{blue}{5}&\color{green}{13}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{green}{17}&\color{blue}{5}&\color{green}{19}\\ \color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{green}{23}&\color{blue}{5}&\color{red}{5}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{29}\\ \color{blue}{7}&\color{green}{31}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{red}{5}&\color{blue}{5}&\color{green}{37}&\color{blue}{7}&\color{red}{3}\\ \color{blue}{3}&\color{green}{41}&\color{blue}{5}&\color{green}{43}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{green}{47}&\color{blue}{5}&\color{red}{7}\\ \color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{53}&\color{blue}{7}&\color{red}{5}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{59}\\ \color{blue}{7}&\color{green}{61}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{red}{5}&\color{blue}{5}&\color{green}{67}&\color{blue}{7}&\color{red}{3}\\ \color{blue}{3}&\color{green}{71}&\color{blue}{5}&\color{green}{73}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{red}{7}&\color{blue}{5}&\color{green}{79} \end{array}$$ $$\begin{array}{cccccccccc} 870&\color{red}{13}&872&\color{red}{3}&874&\color{red}{5}&876&\color{green}{877}&878&\color{red}{3}\\ \color{blue}{3}&\color{green}{881}&\color{blue}{5}&\color{green}{883}&\color{blue}{3}&\color{red}{3}&\color{blue}{3}&\color{green}{887}&\color{blue}{5}&\color{red}{7}\\ \color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{red}{19}&\color{blue}{7}&\color{red}{5}&\color{blue}{13}&\color{red}{3}&\color{blue}{11}&\color{red}{29}\\ \color{blue}{13}&\color{red}{17}&\color{blue}{19}&\color{red}{3}&\color{blue}{17}&\color{red}{5}&\color{blue}{19}&\color{green}{907}&\color{blue}{31}&\color{red}{3}\\ \color{blue}{3}&\color{green}{911}&\color{blue}{5}&\color{red}{11}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{red}{7}&\color{blue}{7}&\color{green}{919}\\ \color{blue}{13}&\color{red}{3}&\color{blue}{3}&\color{red}{13}&\color{blue}{5}&\color{red}{5}&\color{blue}{7}&\color{red}{3}&\color{blue}{17}&\color{green}{929}\\ \color{blue}{11}&\color{red}{7}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{red}{5}&\color{blue}{7}&\color{green}{937}&\color{blue}{19}&\color{red}{3}\\ \color{blue}{3}&\color{green}{941}&\color{blue}{5}&\color{red}{23}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{947}&\color{blue}{7}&\color{red}{13}\\ \color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{953}&\color{blue}{7}&\color{red}{5}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{red}{7}\\ \color{blue}{7}&\color{red}{31}&\color{blue}{43}&\color{red}{3}&\color{blue}{11}&\color{red}{5}&\color{blue}{13}&\color{green}{967}&\color{blue}{31}&\color{red}{3}\\ \color{blue}{3}&\color{green}{971}&\color{blue}{5}&\color{red}{7}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{977}&\color{blue}{7}&\color{red}{11}\\ \color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{green}{983}&\color{blue}{7}&\color{red}{5}&\color{blue}{3}&\color{red}{3}&\color{blue}{5}&\color{red}{23}\\ \color{blue}{7}&\color{green}{991}&\color{blue}{73}&\color{red}{3}&\color{blue}{3}&\color{red}{5}&\color{blue}{5}&\color{green}{997}&\color{blue}{7}&\color{red}{3}\\ \color{blue}{3}&\color{red}{7}&\color{blue}{5}&\color{red}{17}&\color{blue}{7}&\color{red}{3}&\color{blue}{23}&\color{red}{19}&\color{blue}{11}&\color{green}{1009} \end{array}$$
Let $p_1$ be the largest prime less than $900$. $p_1 = 887$. Let $Q_1$ be the list of all the (odd) primes between $900- p_1$ and $1000 - p_1$ $$ 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113 $$ and record which even numbers in $[900,1000]$ are covered by elements of $p_1 + Q_1$. $$ (p_1 + Q_1)\cap [900,1000] = 900, 904, 906, 910, 916, 918, 924, 928, 930, 934, 940, 946, 948, 954, 958, 960, 966, 970, 976, 984, 988, 990, 994, 996, 1000 \text{.} $$ That's one quarter of the problem solved. Now we only need to find representations for a subset of $[902,998]$.
Now let $p_2$ be the largest prime less than $p_1$. $p_2 = 883$. Let $Q_2$ be the list of all the (odd) primes between $902 - p_2$ and $998 - p_2$ $$ 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113 $$ and record which even numbers in $[900,1000]$ are covered by elements of $p_2 + Q_2$. $$ (p_2 + Q_2)\cap [900,1000] = 902^*, 906, 912^*, 914^*, 920^*, 924, 926^*, 930, 936^*, 942^*, 944^*, 950^*, 954, 956^*, 962^*, 966, 972^*, 980^*, 984, 986^*, 990, 992^*, 996 \text{,} $$ where we mark by asterisks those numbers not previously covered. This is 16 more, so we are 41% finished.
Continue until you have covered everything, although covering the last few cases can take a while. You might occasionally switch to directly trying to cover particular unresolved even numbers in the interval. (Happily, the following process can be stopped and continued easily.)