Let's restate the problem for clarity: **Given:** \( 2n + 1 \) and \( 3n + 1 \) are both perfect squares for some integer \( n \). **Prove:** \( n \) is divisible by \( 40 \). *Hint: Show \( n \) is divisible by both \( 5 \) and \( 8 \).* --- Let's proceed step by step. --- ### 1. Set up Let: \[ 2n + 1 = a^2 \] \[ 3n + 1 = b^2 \] for some integers \( a \), \( b \). Then: \[ 2n = a^2 - 1 \implies n = \frac{a^2 - 1}{2} \] \[ 3n = b^2 - 1 \implies n = \frac{b^2 - 1}{3} \] Set them equal: \[ \frac{a^2 - 1}{2} = \frac{b^2 - 1}{3} \] \[ 3(a^2 - 1) = 2(b^2 - 1) \] \[ 3a^2 - 3 = 2b^2 - 2 \] \[ 3a^2 - 2b^2 = 1 \] This is a **Pell-type equation**. --- ### 2. Find all integer solutions to \( 3a^2 - 2b^2 = 1 \) Let's look for integer solutions for \( (a, b) \): Try small integer values: - \( a = 1 \): \( 3(1)^2 - 2b^2 = 1 \Rightarrow 3 - 2b^2 = 1 \Rightarrow b^2 = 1 \Rightarrow b = \pm 1 \) So \( (a, b) = (1, 1) \) is a solution. - \( a = 7 \): \( 3(49) - 2b^2 = 1 \Rightarrow 147 - 2b^2 = 1 \Rightarrow b^2 = 73 \), not integer. Try to find the general solution using the method for such Pell equations. --- ### 3. Find the next solution (using recurrence relations) Consider the general method for solving equations of the form \( x^2 - D y^2 = 1 \). In our case, the equation is not in standard Pell form, but let's try to find a recurrence. Let \( (a_1, b_1) = (1, 1) \). Try to find a recurrence: Suppose \( (a_k, b_k) \) is a solution. Notice that if we can find one solution, we can generate more using: Let’s try to write the equation as: \[ 3a^2 - 2b^2 = 1 \] It's not standard Pell, but let's try anyway. Alternatively, let's try some small values for \( a \): - \( a = 1 \): Already done. - \( a = 5 \): \( 3(25) - 2b^2 = 1 \implies 75 - 2b^2 = 1 \implies b^2 = 37 \), not integer. - \( a = 7 \): already done. - \( a = 13 \): \( 3(169) - 2b^2 = 1 \implies 507 - 2b^2 = 1 \implies b^2 = 253 \), not integer. - \( a = 17 \): \( 3(289) - 2b^2 = 1 \implies 867 - 2b^2 = 1 \implies b^2 = 433 \), not integer. Try \( a = 11 \): \( 3(121) - 2b^2 = 1 \implies 363 - 2b^2 = 1 \implies b^2 = 181 \), not integer. Try to find a pattern. Alternatively, try solving for \( a \) in terms of \( b \): \[ 3a^2 = 2b^2 + 1 \implies a^2 = \frac{2b^2 + 1}{3} \] So \( 2b^2 + 1 \) must be divisible by 3, so \( b^2 \equiv 1 \pmod{3} \implies b \equiv \pm1 \pmod{3} \). Let’s try \( b = 1 \): \( a^2 = 1 \), so \( a = 1 \) (already found). Try \( b = 2 \): \( a^2 = \frac{2 \cdot 4 + 1}{3} = \frac{8 + 1}{3} = 3 \), not a square. Try \( b = 4 \): \( a^2 = \frac{2 \cdot 16 + 1}{3} = \frac{32 + 1}{3} = 11 \), not a square. Try \( b = 5 \): \( a^2 = \frac{2 \cdot 25 + 1}{3} = \frac{50 + 1}{3} = 17 \), not a square. Try \( b = 7 \): \( a^2 = \frac{2 \cdot 49 + 1}{3} = \frac{98 + 1}{3} = 33 \), not a square. Try \( b = 11 \): \( a^2 = \frac{2 \cdot 121 + 1}{3} = \frac{242 + 1}{3} = 81 \), which is \( 9^2 \). So **\( b = 11 \), \( a = 9 \) is another solution.** So, with \( a = 9 \), \( b = 11 \): Let’s compute \( n \): \[ n = \frac{a^2 - 1}{2} = \frac{81 - 1}{2} = 40 \] \[ 2n + 1 = 2 \times 40 + 1 = 81 = 9^2 \] \[ 3n + 1 = 3 \times 40 + 1 = 121 = 11^2 \] **So \( n = 40 \) works.** --- ### 4. Find the general solution Note the pattern: \( (a_1, b_1) = (1, 1) \), \( (a_2, b_2) = (9, 11) \). Let's see if there's a recurrence. Looking at the equations: Given \( (a_k, b_k) \), is there a recurrence formula? For Pell equations, often: \[ (a_{k+1} + b_{k+1}\sqrt{d}) = (a_1 + b_1\sqrt{d})^{k} \] But our equation is \( 3a^2 - 2b^2 = 1 \). Alternatively, let's see if we can see a pattern in \( n \): - First non-trivial solution: \( n = 40 \). - Try to find the next solution. Try the next \( b \), which should be \( b > 11 \), \( b \equiv 1 \pmod{3} \). Try \( b = 23 \): \[ a^2 = \frac{2 \cdot 529 + 1}{3} = \frac{1058 + 1}{3} = 353 \] Not a square. Try \( b = 29 \): \[ a^2 = \frac{2 \cdot 841 + 1}{3} = \frac{1682 + 1}{3} = 561 \] Not a square. Try \( b = 41 \): \[ a^2 = \frac{2 \cdot 1681 + 1}{3} = \frac{3362 + 1}{3} = 1121 \] Not a square. Try \( b = 71 \): \[ a^2 = \frac{2 \cdot 5041 + 1}{3} = \frac{10082 + 1}{3} = 3361 \] Not a square. Try \( b = 61 \): \[ a^2 = \frac{2 \cdot 3721 + 1}{3} = \frac{7442 + 1}{3} = 2481 \] Not a square. Alternatively, let's try to find the recurrence between the solutions. From the two solutions: - \( (a_1, b_1) = (1, 1) \) - \( (a_2, b_2) = (9, 11) \) Let’s try to find the recurrence: Let’s try to express \( (a_{k+1}, b_{k+1}) \) in terms of \( (a_k, b_k) \): Suppose \( a_{k+1} = p a_k + q b_k \) and similarly for \( b_{k+1} = r a_k + s b_k \). Plugging in the first two solutions: - \( a_2 = 9 = p \cdot 1 + q \cdot 1 = p + q \) - \( b_2 = 11 = r \cdot 1 + s \cdot 1 = r + s \) Now, let's try to find the next solution explicitly: Let’s try to use the standard method for quadratic forms. Alternatively, since the only solution for \( n \) in positive integers is \( n = 40 \), let's check if divisibility by 5 and 8 is always guaranteed. --- ### 5. Show divisibility by 5 and 8 Recall: \[ n = \frac{a^2 - 1}{2} \] and \[ n = \frac{b^2 - 1}{3} \] Set \[ a^2 \equiv 1 \pmod{2} \implies a \text{ is odd} \] Similarly, \[ b^2 \equiv 1 \pmod{3} \implies b \equiv \pm 1 \pmod{3} \] Let’s analyze \( n \pmod{5} \): Suppose \( n = \frac{a^2 - 1}{2} \), and \( 3n + 1 = b^2 \). So, \[ 3n + 1 = b^2 \implies 3 \cdot \frac{a^2 - 1}{2} + 1 = b^2 \implies \frac{3a^2 - 3}{2} + 1 = b^2 \implies \frac{3a^2 - 1}{2} = b^2 \implies 3a^2 - 1 = 2b^2 \implies 3a^2 - 2b^2 = 1 \] Now, consider \( n \mod 5 \): Let’s see what possible values \( a^2 \) can take mod 5: - \( 0^2 = 0 \) - \( 1^2 = 1 \) - \( 2^2 = 4 \) - \( 3^2 = 9 = 4 \) - \( 4^2 = 16 = 1 \) So, \( a^2 \equiv 0, 1, 4 \pmod{5} \) Now, \( n = \frac{a^2 - 1}{2} \), so \( a^2 \equiv 1 \pmod{2} \implies a \) odd. Try \( a^2 \equiv 0 \pmod{5} \implies a^2 = 0 \), \( n = \frac{-1}{2} \), not integer. Try \( a^2 \equiv 1 \pmod{5} \implies n \equiv 0 \pmod{5} \). Try \( a^2 \equiv 4 \pmod{5} \implies n \equiv \frac{3}{2} \pmod{5} \), not integer. So only \( a^2 \equiv 1 \pmod{5} \) yields integer \( n \). Thus, \( n \equiv 0 \pmod{5} \), so **\( n \) is divisible by 5**. Now, check \( n \mod 8 \): Possible odd squares mod 8: - \( 1^2 = 1 \) - \( 3^2 = 9 = 1 \) - \( 5^2 = 25 = 1 \) - \( 7^2 = 49 = 1 \) So for odd \( a \), \( a^2 \equiv 1 \pmod{8} \). Thus, \[ n = \frac{a^2 - 1}{2} = \frac{1 - 1}{2} = 0 \pmod{8} \] So **\( n \) is divisible by 8**. Thus, **\( n \) is divisible by 5 and 8**, and hence by 40. --- ### 6. **Conclusion** **If \( 2n+1 \) and \( 3n+1 \) are both perfect squares, then \( n \) is divisible by 40.** --- ## **Summary of Proof** Let \(