2025/033) Let a,b,c be the consecutive sandwiched integers between any cousin primes. Is it true that either $5∣(a+b+c)$ or $5| (a^2+b^2+c^2)$ ?

 Let x,y be cousin primes. We have x,a,b,c,y are 5 numbers.

now a = b-1 and c = b +1

so We have $a+b+c = 3b$

$a^2+b^2+c^2 = (b-1)^2 + b^2 + (b+1)^2 = 3b^2 + 2$

x is a prime so x cannot be multiple of 5 or b cannot be of the form 5k+2

y is a prime so y cannot be multiple of 5 or b cannot be of the form 5k +3

If b is of the form 5k $a+b+c$ is divisible by 5

If b is of the form 5k+ 1  $a^2+b^2+c^2 = 3(5k+1)^2 + 2 = 3(25k^2+10k+1) + 2 = 75k^2+ 30k + 5 = 5(15k^2+ 6k+1)$ multiple of 5

 If b is of the form 5k+ 4  $a^2+b^2+c^2 = 3(5k+4)^2 + 2 = 3(25k^2+40k+16) + 2$

$= 75k^2+ 120k + 50 = 5(15k^2+24k+10)$ multiple of 5

 So either of them is multiple of 5