Good morning from Cambridge! I hope everyone had a great Easter weekend I spent mine cooking brunch. Now that everyone is back to work, take a short break and solve our Problem of the Week. Like usual, I posted a challenge problem to our Facebook, Twitter, and Google+ pages, and it can be found right here:

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A proof! This one doesn't look too tricky. I've uploaded my solution (hint: use modular numbers!), and it can be found below...

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What I did was make a chart. The remainders when we divide f(x) by 5 must be 0, 1, 2, 3, or 4. Then I separated the components of f(x) x obviously has a remainder of 0, 1, 2, 3, or 4 when divided by 5. Then x^2 was a little more complicated, but it was easy to find its remainders, and it is cyclical as well, repeating when x=5. Then I added the remainders and 1, and when converted to mod5, we can see that no integer x value will ever make the equation equal to 0.
Are you confused about a step I took? Did you prove the statement in a different way? Leave a comment and I'll try to reply! Thanks for playing, have a great week!
Tori
Slightly different way:
ReplyDeleteSuppose 5  xx + x + 1, then xx + x + 1 = 0 (mod 5), and xx + x = x(x+1) = 4 (mod 5).
4 (mod 5) decomposes under multiplication into only: 2*2, and 4*1, neither of which has the form x(x+1), hence there is no integer satisfying the statement 5  xx + x + 1.
Best, sea.
Nice work! It's always good to see how many different methods exist for a proof. Tori
DeleteMinor edit, I missed 3*3 in that list.
ReplyDeleteSimilar to the published soln but without the modular stuff is to say that all numbers are of one of the forms 5n, 5n+1, 5n+2, 5n+3, 5n+4 and then factorise the answers.
ReplyDeleteEg
5n+2: (5n+2)²+(5n+2)+1 = 5(5n²+5n+1)+2
A number is only divisible by 5 if its last digit is 0 or 5. X*X + X= X(X+1) which is clearly even and so F(X) is odd. It is simple to show the the product of consecutive integers, X(X+1) results in integers that only end with 0, 2 or 6; never 4. So F(X) can never end in 5.
ReplyDelete