1) t=5(2n+1) where integer n≥0.
t2=25(4n2+4n+1)=100n2+100n+25=100n(n+1)+25, which means that we have a multiple of 100 added to 25. So the square ends in 25.
2) The integers can be written n-1, n, n+1. If n is even then n-1 and n+1 are both odd so their product is odd. When multiplied by n, however, we always get an even number (it's the times 2 table). If n is odd the other two integers must be even so their product is even and mnp is even.
3) n2-n=n(n-1), which is the product of an odd and even number (because they're consecutive), which is always even.
4) In the following m and n are arbitrary natural numbers. There are four cases for integers p and q: a) p and q are both even: p=2m, q=2n, product is 4mn which is even; b) p=2m, q=2n+1, pq=4mn+2m=2(2mn+m) which is even; c) p=2m+1, q=2n, pq=4mn+2n=2(2mn+n) which is even; d) p=2m+1, p=2n+1, pq=4mn+2m+2n+1=2(2mn+m+n)+1 which is odd. In a), b), c) at least one of the integers is even. In case d) both are odd, so the product is only even if at least one of the integers is even.