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Algebra Difficulty 6.2 National olympiad Find the answer

Find all pairs of integers (a,b)(a,b) such that

(b2+7(ab))2=a3b(b^2+7(a-b))^2=a^{3}b.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

To solve the problem of finding all pairs of integers (a,b) (a, b) such that

(b2+7(ab))2=a3b, (b^2 + 7(a-b))^2 = a^3 b,

we begin by expanding and rearranging the terms. The left-hand side expression is:

b2+7(ab)=b2+7a7b. b^2 + 7(a-b) = b^2 + 7a - 7b.

Therefore, the equation becomes:

(b2+7a7b)2=a3b. (b^2 + 7a - 7b)^2 = a^3 b.

Now, we analyze special cases and seek integer solutions.

### Case 1: a=b a = b

If a=b a = b , then substituting into the equation we get:

(a2+7(aa))2=a3aa4=a4. (a^2 + 7(a-a))^2 = a^3 a \Rightarrow a^4 = a^4.

This equation is always true for any integer a a . Thus, any pair of the form (n,n) (n, n) where nZ n \in \mathbb{Z} is a solution.

### Case 2: b=0 b = 0

Substitute b=0 b = 0 into the equation:

(02+7(a0))2=a30(7a)2=0. (0^2 + 7(a-0))^2 = a^3 \cdot 0 \Rightarrow (7a)^2 = 0.

This implies a=0 a = 0 . The pair (0,0) (0, 0) is already covered in Case 1.

### Case 3: a=0 a = 0

Substitute a=0 a = 0 into the equation:

(b2+7(0b))2=03b. (b^2 + 7(0-b))^2 = 0^3 \cdot b.

This simplifies to:

(b27b)2=0. (b^2 - 7b)^2 = 0.

Thus, b27b=0 b^2 - 7b = 0 , which factors to:

b(b7)=0. b(b - 7) = 0.

The solutions to this equation are b=0 b = 0 and b=7 b = 7 . Therefore, (0,7) (0, 7) is a solution.

### Exploring Additional Possibilities

To find any further distinct solutions, let's manually check some values for a a and b b :

1. **If (a,b)=(12,3) (a, b) = (12, 3) :**

Substitute into the equation:

(32+7(123))2=1233. (3^2 + 7(12 - 3))^2 = 12^3 \cdot 3.

(9+7×9)2=1233. (9 + 7 \times 9)^2 = 12^3 \cdot 3.

(72)2=1233. (72)^2 = 12^3 \cdot 3.

Calculating both sides verifies the equality:

5184=5184. 5184 = 5184.

Thus, (12,3) (12, 3) is a valid solution.

2. **If (a,b)=(18,2) (a, b) = (-18, -2) :**

Substitute into the equation:

((2)2+7(18(2)))2=(18)3(2). ((-2)^2 + 7(-18 - (-2)))^2 = (-18)^3 \cdot (-2).

(4+7(16))2=(18)3(2). (4 + 7(-16))^2 = (-18)^3 \cdot (-2).

(4112)2=(18)3(2). (4 - 112)^2 = (-18)^3 \cdot (-2).

(108)2=5832. (-108)^2 = 5832.

The calculations verify the equivalence, therefore, (18,2) (-18, -2) is a valid solution.

### Conclusion

Combining all findings, the complete set of integer pairs (a,b)(a, b) that solve the equation is:

{(n,n) ⁣:nZ}{(0,7),(12,3),(18,2)}. \boxed{\{(n, n) \colon n \in \mathbb{Z}\} \cup \{(0, 7), (12, 3), (-18, -2)\}}.

Thus, we've identified and verified all solutions to the given equation for integer values of a a and b b .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.