Since u must be positive we have 3 2 x = 1 or 3 2 x = 1

A4919.

Solution by the Missouri State University Problem Solving Group Department of Mathematics

Missouri State University 901 S. National Avenue Springfield, MO 65897 Q1: We have

(3³)^x+ (2³)^x = 2·(2·3²)^x. Dividing both sides by (2·3²)^x and simplifying, we have

3

2

x

+

3

2

x−2

= 2.

Letting u= (3/2)^x we obtain

u+u⁻² = 2, hence u³−2u²+ 1 = 0.

Now u= 1 is a root and factoring gives

(u−1)(u²−u−1) = 0, so the roots are 1 and (1±√

5)/2. Since u must be positive we have

3

2

x

= 1 or

3

2

x

= 1 +√ 5 2 , giving

x= 0 or x= ln (1 +√ 5)/2 ln(3/2) .

One readily verifies that these do satisfy the original equation.

Q2: Suppose that y is odd, sayy = 2k+ 1. Then

x²+ 26455≡x² ≡2^2k+1 ≡2(−1)^k (mod 5).

However ±2 is not a square modulo 5, so this is impossible.

If y= 2k, then

2^k+x

2^k−x

= 2^2k−x² = 26455.

1

The positive divisors of 26455 are

d= 1,5,11,13,37,55,65,143,185,407,481,715,2035,2405,5291, and 26455.

Therefore

2^k+1 = 2^k+x

+ 2^k−x

=d+26455 d .

The only values for which d+ 26455/d is a power of 2 are d = 13,2035 and hence 2^k+1 = 2048, so k+ 1 = 11 andy = 20. This givesx² = 2²⁰−26455 = 1022121 and hence x= 1011.

Q3: Since our sequence is geometric, we have 10 +x

4000 +x =r⁶ for some rational number r. This gives

10 +x=λa⁶ and 4000 +x=λb⁶ for integers a, b, and λ. But then

3990 = (4000 +x)−(10 +x) =λ b⁶−a⁶ .

The only positive values of b⁶−a⁶ less than or equal to 3990 are 2⁶−1⁶,3⁶− 1,3⁶−2⁶,and 4⁶−3⁶. The only one of these that is a factor of 3990 is 3⁶−2⁶. Therefore, we must havea= 2, b= 3,andλ= 6. This gives 4000+x= 4374, so fourth term in the sequence is therefore

4374

2

3 3

= 1296.

2