How Many Pairs of Positive Integers (x, y) Satisfy the Equation 3x^4y 120?

To determine the pairs of positive integers (x, y) that satisfy the equation 3x^4y 120, we start by rearranging the equation to isolate y:

Step 1: Rearrange the Equation

By rearranging the equation, we obtain:

4y 120 - 3x

Step 2: Ensure y is a Positive Integer

For y to be a positive integer, 120 - 3x must be a positive multiple of 4. This leads us to the following conditions:

120 - 3x 0, which implies x 40.

120 - 3x ≡ 0 mod 4.

Step 3: Simplify the Modulo Condition

Calculating 120 mod 4, we find:

120 ≡ 0 mod 4

Now we compute 3x mod 4. Since 3 ≡ -1 mod 4, we have:

3x ≡ -x mod 4

Thus, for 120 - 3x ≡ 0 mod 4, we need:

-x ≡ 0 mod 4 which implies x ≡ 0 mod 4.

This means x must be a multiple of 4. Let x 4k for some positive integer k.

Step 4: Substitute x 4k into the Equation

Substituting x 4k back into the equation gives:

120 - 3(4k) 4y

120 - 12k 4y

4y 120 - 12k which simplifies to:

y 30 - 3k

Step 5: Ensure Both x and y are Positive Integers

We need both x 4k and y 30 - 3k to be positive integers:

4k 0, which implies k 0.

30 - 3k 0 which implies 30 3k or k 10.

Thus, k must be a positive integer such that 1 ≤ k ≤ 10. The possible values for k are: 1, 2, 3, 4, 5, 6, 7, 8, 9.

This gives us 9 possible values for k.

Step 6: Calculate Pairs (x, y)

Now we can calculate the pairs (x, y) for each k:

k 1 → x 4(1) 4, y 30 - 3(1) 27 rarr; (4, 27) k 2 → x 4(2) 8, y 30 - 3(2) 24 rarr; (8, 24) k 3 → x 4(3) 12, y 30 - 3(3) 21 rarr; (12, 21) k 4 → x 4(4) 16, y 30 - 3(4) 18 rarr; (16, 18) k 5 → x 4(5) 20, y 30 - 3(5) 15 rarr; (20, 15) k 6 → x 4(6) 24, y 30 - 3(6) 12 rarr; (24, 12) k 7 → x 4(7) 28, y 30 - 3(7) 9 rarr; (28, 9) k 8 → x 4(8) 32, y 30 - 3(8) 6 rarr; (32, 6) k 9 → x 4(9) 36, y 30 - 3(9) 3 rarr; (36, 3)

In total, there are 9 pairs of positive integers (x, y) that satisfy the equation 3x^4y 120.