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Question 1 from 25th British Mathematical Olympiad, 1988

Hey,

I have been attempting this question for the past hour, and I have made no progress:

Find all integers a, b, and c for which

(x - a)(x - 10) + 1 = (x + b)(x + c) for all x

(Question 1 from 25th British Mathematical Olympiad, 1988)

Could you give me a clue to start attempting this question? Thank you.

Reply 1

DFranklin

What have you tried so far?

Reply 2

mqb2766

Original post by entrepreneur_13

Have you tried expanding and equating coeffs?

Reply 3

entrepreneur_13

I have expanded the question and I got:

-ax - 10x + 1 = bx + cx + bc

But how do you equate coefficient in this case?

Reply 4

DFranklin

Original post by entrepreneur_13

I have expanded the question and I got:

-ax - 10x + 1 = bx + cx + bc

But how do you equate coefficient in this case?

Just like for any other identity:

https://en.wikipedia.org/wiki/Equating_coefficients

Reply 5

DFranklin

Original post by entrepreneur_13

I have expanded the question and I got:

-ax - 10x + 1 = bx + cx + bc

But how do you equate coefficient in this case?

Just actually looked at your expansion and noticed it's wrong. Without being snarky (and I do understand we all make mistakes from time to time) - if you can't expand out simple expressions like this reliably, BMO questions might be a bit of a reach for you right now.

Reply 6

entrepreneur_13

Oops sorry. I messed up while copying what I wrote down. I think I understand what to do now:

-ax - 10x + 10a + 1 = bx + cx + bc
x (-a - 10) + 10a + 1 = x(b+c) + bc
-a -10 = b+c
10a+1 = bc

I will try to solve it from here now (hopefully this is right).

Reply 7

entrepreneur_13

I eliminated a and factorised the resulting equation to get (b+10)(c+10) = 1.
I solved for b and c, and substituted into -a - 10 = b + c to find the value for a.

My final solutions are:
b = -9, c = -9, a = -8, x = 10
b = -11, c = -11, a = 12, x = 0

Reply 8

ghostwalker

Original post by entrepreneur_13

My final solutions are:
b = -9, c = -9, a = -8, x = 10
b = -11, c = -11, a = 12, x = 0

In red.

Agree with your b,c.

There's no restriction on x.

Reply 9

davros

Original post by entrepreneur_13

As an alternative to comparing coefficients you could start by setting x = 10 (since you're told the relation holds for all x) - this simplifies the LHS and tells you something about the product of 2 integers on the RHS :smile:

Reply 10

DFranklin

Original post by davros

Very nice. (PRSOM)...

Reply 11

entrepreneur_13

Here is my solution:

(x-a)(x-10) + 1 = (x+b)(x+c)
x^2 - ax - 10x + 10a + 1 = x^2 + bx + cx + bc
x^2 + x(-a-10) + 10a + 1 = x^2 + x(b+c) + bc
-a-10 = b+c
-a = b + c + 10
a = - b - c - 10

10a + 1 = bc

10(-b-c-10) + 1 = bc

-10b - 10c - 99 = bc

bc + 10b + 10c = 99
(b + 10)(c + 10) = 1
b + 10 = 1; c + 10 = 1; hence, b = -9, c = -9, a = 8

b + 10 = -1; c + 10 = -1; hence, b = -11, c = -11, a = 12

Reply 12

mqb2766

Original post by entrepreneur_13

That looks good. I take it you read davros hint about another way to start off instead of equating coeffs you can sub x=10 to get
1 = (10+b)(10+c)
directly.

Also its worth thinking about it as a factorized quadratic with unity quadratic coeff. The left one has roots at 10 and a which is then shifted up by 1, The right at -b and -c. So -b anc -c must lie inside 10..a or a..10. Also as the +1 vertical translation is small, a must be close to 10 for all the solutions/roots to be integers, which would have caught your previous sign error. This is not a proof by any stretch, but it helps to visualize the problem/verify solutions. By being a bit more precise with the vertical translation and the spacing between the roots, it could be made solid (thinking about the discriminant), but its probably more complex solution overall.

The bmo problem i referred to yesterday was the other one, rather than this one.

(edited 1 year ago)

Reply 13

davros

Original post by entrepreneur_13

As an alternative, setting x= 10 immediately gives you "product of 2 integers = 1" which restricts possible values of b and c very quickly. Similarly, setting x = a gives you exactly the same sort of condition, forcing values for a depending on your choice of b and c :smile: