Number theory

Number Theory Level pending

How many ordered pairs of integers $(a,b)$ are there such that

$a^4,b^2+1,1$ are in an arithmetic progression?

The answer is 2.

1 solution

Akash Shukla
Jun 26, 2016

$2(b^2+1) = a^4 + 1$ ,

$2b^2 = a^4-1$ ,

$b^2 = \dfrac{a^4 -1}{2}$ ,

now b is integer, then last digit of $a^4$ must be odd,i.e. $(1,5,9)$ . As the last digit of $a^4$ can't be $9$ . So last digit of $a^4$ is $1,5$ .So last digit of $b^2$ are $0,5,4,7$ .As it is perfect square, so $4$ and $7$ are neglected. So last digit of $b^2$ are $0,5$ . So last digit of $b$ is $0,5$ .So $A=0,B=5$ . Therefore $A+B-1 = \boxed{4}$

Moderator note:

This solution is incomplete as it only provides a necessary condition. It does not show that the condition is sufficient.

The only integer solutions to the equation is $\pm 1, 0$ .

Your solution shows a necessary condition. It has not proven that the condition is sufficient.

I have edited this problem for clarity.

Calvin Lin Staff - 4 years, 11 months ago

Ther exists no such integer a,b

Kushal Bose - 4 years, 11 months ago

Number theory

The answer is 2.

1 solution

Moderator note:

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