Problem:
For integers a,b,c, and d, let f(x)=x2+ax+b and g(x)=x2+cx+d. Find the number of ordered triples (a,b,c) of integers with absolute values not exceeding 10 for which there is an integer d such that g(f(2))=g(f(4))=0.
Solution:
Notice that since we have ordered triples (a,b,c) such that g(f(2))=g(f(4))=0,
that we can notice that for these given triples g(f(2))βg(f(4))=0β0=0
We then notice that a valid d can be found given that the conditions above are met.\
Let's evaluate g(f(4))βg(f(2))=0
(f(4))=f(4)2+c(f(4))+d
and
g(f(2))=f(2)2+c(f(2))+d
Subtracting these yields
g(f(4))βg(f(2))=f(4)2βf(2)2+c(f(4)βf(2))=0
Using difference of squares, we can factor this as
(f(4)βf(2))(f(4)+f(2)+c)=(12+2a)(20+c+6a+2b)
Let's remember that this evaluates to zero so either of these will have to evaluate to 0.
Starting with the first part (12+2a), this evaluates to zero when a=β6,
and we have 21 choices for each of b and c as they range from β10 to 10 inclusive, yielding 212=441 triples.
Now, we have that
20+c+6a+2b=0βc=β20β6aβ2b
but have to also remember that the absolute values range from β10 to 10
Thus, let's use the boundaries of c to rewrite c=β20β6aβ2b as
β10β€β20β6aβ2bβ€10
and rearranging we get
β30β€6a+2bβ€β10
From here, we just bash over possible values of a that range from -8 to 1, excluding -6 as we've calculated it already.
The results are as follows
When a=β8, there are 2 values of b, 9 and 10
Then when a=β7, there are 5 values of b, 6,7,...,10
Then when a=β5, there are 11 values, namely 0,1,...,10
When a=β4, there are 11, namely β3,...,7
When a=β3, there are 11 again, namely β6,...,4
When a=β2, there are 11, β9,...,1
When a=β1, there are 9, namely β10,...,β2
When a=0, there are 6, namely β10,...,β5
and Lastly
when a=1, there are 3, namely β10,...,β8
Adding these up yields
2+5+11+11+11+11+9+6+3=69
and we add the 212 we found in the first case, yielding 441+69=510β triples
old solution below`
For a given ordered triple (a,b,c), the value of g(f(4))βg(f(2)) is uniquely determined, and a value of d can be found to give g(f(2)) any prescribed integer value. Hence the required condition can be satisfied provided that a,b, and c are chosen so that
0β=g(f(4))βg(f(2))=(f(4))2β(f(2))2+c(f(4)βf(2))=(f(4)βf(2))(f(4)+f(2)+c)=(12+2a)(20+6a+2b+c)β
First suppose that 12+2a=0, so a=β6. In this case there are 21 choices for each of b and c with β10β€bβ€10 and β10β€cβ€10, so this case accounts for 212=441 ordered triples.
Next suppose that aξ =β6 and 20+6a+2b+c=0, so c=β6aβ2bβ20. Because β10β€cβ€10, it follows that β30β€6a+2bβ€β10, and because β10β€bβ€10, it follows that β8β€aβ€1. Then β15β3aβ€bβ€β5β3a. The number of ordered triples for various values of a are presented in the following table.
aβ8β7β6β5β4β3β2β101Totalβb{9,10}{6,7,8,9,10}{β10,β¦,10}{0,β¦,10}{β3,β¦,7}{β6,β¦,4}{β9,β¦,1}{β10,β¦,β2}{β10,β¦,β5}{β10,β9,β8}βcβ6aβ2bβ20β6aβ2bβ20{β10,β¦,10}β6aβ2bβ20β6aβ2bβ20β6aβ2bβ20β6aβ2bβ20β6aβ2bβ20β6aβ2bβ20β6aβ2bβ20βtriples2544111111111963510ββ
The total number of ordered triples that satisfy the required condition is 510β.
The problems on this page are the property of the MAA's American Mathematics Competitions