Hypotenuse of largest Pythagorean triangle whose sides differ by 1 and hypotenuse < 2^31

Number Theory Level 3

Find the largest possible right triangle (in terms of area) with

  • Integer side lengths,
  • The two shorter sides differ by 1, and
  • The hypotenuse has a length of less than 2 31 2^{31} .

Submit your answer as the hypotenuse of this triangle.


The answer is 1311738121.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Hypotenuses of Pythagorean triangle whose sides differ by 1 are every other elements of the Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2), starting with the element whose value is 5. The sequence is A000129 .

Here is a sufficient subsequence of A000129 with the bit length of number as a unsigned binary integer. 5 3 29 5 169 8 985 10 5741 13 33461 16 195025 18 1136689 21 6625109 23 38613965 26 225058681 28 1311738121 31 7645370045 33 \begin{array}{rr} 5 & 3 \\ 29 & 5 \\ 169 & 8 \\ 985 & 10 \\ 5741 & 13 \\ 33461 & 16 \\ 195025 & 18 \\ 1136689 & 21 \\ 6625109 & 23 \\ 38613965 & 26 \\ 225058681 & 28 \\ 1311738121 & 31 \\ 7645370045 & 33 \\ \end{array}

It can be seen easily that the answer is 1311738121 1311738121 . The other two sides are { ⌊ 1311738121 2 ⌋ , ⌈ 1311738121 2 ⌉ } \left\{\left\lfloor \frac{1311738121}{\sqrt{2}}\right\rfloor ,\left\lceil \frac{1311738121}{\sqrt{2}}\right\rceil \right\} or { 927538920 , 927538921 } \{927538920,927538921\} and 92753892 0 2 + 92753892 1 2 \sqrt{927538920^2+927538921^2} is 1311738121 1311738121 .

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The OEIS also says that consecutive Pell numbers P n , P n + 1 P_n, P_{n+1} can generate pythagorean triples with sides differing by 1 using the general formula for pythagorean triples

{ a = P n 2 − P n + 1 2 b = 2 P n P n + 1 c = P n 2 + P n + 1 2 \begin{cases} a = P_n^2 - P_{n+1}^2 \\ b = 2 P_n P_{n+1} \\ c = P_n^2 + P_{n+1}^2 \end{cases}

So, I found 13860 and 33461, which also give 1311738121.

Henry U - 2 years, 5 months ago

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Good work!

A Former Brilliant Member - 2 years, 5 months ago

If you're going to use OEIS, you might as well use A001653 , "numbers k k such that 2 k 2 − 1 2k^2-1 is a square." These are precisely the desired hypotenuses: c 2 = a 2 + ( a + 1 ) 2 ⟺ 2 c 2 = 4 a 2 + 4 a + 2 ⟺ 2 c 2 − 1 = ( 2 a + 1 ) 2 \begin{aligned} && c^2 &= a^2 + (a+1)^2 \\ &\iff & 2c^2 &= 4a^2 + 4a + 2 \\ &\iff & 2c^2 - 1 &= (2a + 1)^2 \end{aligned} This is equivalent to 2 c 2 − 1 2c^2 - 1 being an odd square (note that 2 c 2 − 1 2c^2 - 1 can never be an even square).

Jordan Cahn - 2 years, 5 months ago

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