yes, you get the remainder 1 from a of them and 0 from (5-a) of them since 3|(5-a) of them, I got this partThere are 5 squares of which [imath]5-a[/imath] are divisible by 3 and [imath]a[/imath] of them have remainder 1 when divided by one. [imath]1\cdot a[/imath] is the same as just [imath]a[/imath]. From that we know that [imath]a[/imath] has remainder of 1 when divided by 3, and since [imath]0 \leq a \leq 5[/imath] we know that [imath]a[/imath] can be either 1 or 4.
Does this make sense?
Correction: since we are talking about primes "divisible by 3" means "equal to 3" of course.There are 5 squares of which [imath]5-a[/imath] are divisible by 3 and [imath]a[/imath] of them have remainder 1 when divided by one. [imath]1\cdot a[/imath] is the same as just [imath]a[/imath]. From that we know that [imath]a[/imath] has remainder of 1 when divided by 3, and since [imath]0 \leq a \leq 5[/imath] we know that [imath]a[/imath] can be either 1 or 4.
Does this make sense?
Does the rest make sense too?yes, you get the remainder 1 from a of them and 0 from (5-a) of them since 3|(5-a) of them, I got this part
yeah I got it not, thanks for your helpDoes the rest make sense too?
To answer your question about the significance of 3 and 8: there is none, i.e., whatever works