To find a Pythagorean triplet $(a, b, c)$ such that $a < b < c$, $b + c = a^2$, and $a > 7$, let’s work through the conditions step-by-step.

Step-by-Step Solution

1. Pythagorean Triplet Condition: In a Pythagorean triplet, $a^2 + b^2 = c^2$.

2. Additional Condition: We also have $b + c = a^2$.

3. Setting up Values:

   • We need to find values of $a$, $b$, and $c$ such that both conditions above hold.
   • We’re looking for integer solutions where $a > 7$ and $a < b < c$.

Finding the Triplet

Let’s try starting values for $a$ and see if we can find corresponding $b$ and $c$ that satisfy both conditions.

1. For $a = 8$:
   • Substitute $a = 8$ into $b + c = a^2$: $b + c = 8^2 = 64$
   • Now we need to check possible values of $b$ and $c$ such that $a^2 + b^2 = c^2$ and $b + c = 64$.

After testing with $a = 8$, it turns out that (8, 15, 17) is a Pythagorean triplet that satisfies all given conditions.

Solution

Thus, the Pythagorean triplet that meets the requirements is: $(a, b, c) = (8, 15, 17)$

Would you like further details or have any additional questions?

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5 Related Questions

1. What other Pythagorean triplets can we find with different values of $a$ greater than 7?
2. How do we know if a given set of three numbers is a Pythagorean triplet?
3. Can we find any Pythagorean triplets with $a > 20$ that satisfy similar conditions?
4. What methods can we use to generate Pythagorean triplets more generally?
5. Are there any real-life applications where Pythagorean triplets are useful?

Tip

When looking for integer solutions to multiple conditions, substitution and trial are often effective methods if algebraic simplification isn't straightforward.