Now i realize i could use the division algorithm and get to the correct answer, but i was looking for a different way. However this is my first proofing class was wondering if this is valid or not.
5|a if a has a 0 or a 5 in the one's position. So 5|n^2+2 if n^2 has a 3 or an 8 in the ones position. When multiplying 2 numbers the one's position of the product is only affected by the one's position of the multiplicand and the multiplier. This is shown by abc*def=f(abc)+10e(abc)+100d(abc) where a and d are single digit integers in the hundred's place of the number of the multiplicand and multiplier, b and e are single digit integers in the ten's place of the multiplicand and multiplier, and c and f are single digit integers in the one's position of the multiplicand and multiplier. From this result we see that only f(abc) will have an impact on the one's position of the product because any integer multiplied by a factor of ten will have a 0 in the one's position. Also f*(abc)= f*c+f*(10b)+f*(100a) so only f*c will affect the one's position of abc*def. Any integer squared will have the same value for the number in the one's position. We can see that 1^2=1 2^2=4 3^2=9 4^16 5^2=25 6^2=36 7^2=49 8^2=64 9^2=81 0^2=0. Since all of the squares of single digit integers do not end in 3 or 8 in the one's position 5|n^2 +2 is never true.
Was my idea on how to approach it differently. This is my 3rd proof to date.
After Diffe Q and Linear you start getting into proofing, which like people told Luis is very different from calc... its not impossible its just there are no set algorithms to follow that will lead you to the answer. It is fun because its a whole new type of puzzle, i found calc to be pretty easy and Diffe Q is pretty much the same, wish we would have touched PDE's instead of waiting until senior year to introduce this.