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Find all right-angled triangles whose hypotenuse has length $2^{2015.5}$ and whose other two sides have integral lengths.

My attempt: As $2^{2015.5}=2^{2015} \sqrt2$, the other lengths of the triangle can be $(2^{2015},2^{2015})$. Are there any other possible cases?

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    $\begingroup$ Hint: You want to solve $x^2+y^2=2^{4031}$ for $x,y \in \mathbb N$. Look at the Gaussian integers. $\endgroup$ – lhf Sep 20 '16 at 15:33
  • $\begingroup$ Please tell me how to solve that equation sir because I don't know how to solve it $\endgroup$ – sai saandeep Sep 20 '16 at 15:41
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A natural attempt is to work in the ring $ \mathbf Z[i] $ in order to exploit its factoriality. $ 2 $ factors as $ (1+i)(1-i) $ in this ring, and our equality becomes

$$ (x + yi)(x - yi) = (1 + i)^n (1 - i)^n = i^n (1 - i)^{2n}$$

where $ n = 4031 $. The right hand side is a prime factorization in the factorial ring $ \mathbf Z[i] $, and the quantities on the left hand side are conjugates. This tells us that the solutions are of the form

$$ x + yi = \varepsilon (1 - i)^{n} $$

where $ \varepsilon = \pm 1, \pm i $. Accounting for the signs, there is only one solution such that both $ x $ and $ y $ are positive. Since $ x = y = 2^{2015} $ is clearly a solution, we are done.


An alternative approach: We proceed by induction.

Claim: Let $ n = 2k + 1 $ be odd. Then, the equation $ x^2 + y^2 = 2^n $ has a unique solution in the positive integers, given by $ x = y = 2^k $.

Proof. The claim is certainly true for $ k = 0 $. Assume that it is true for $ k $, and let $ m = k+1 $. We want to show that the equation $ x^2 + y^2 = 2^{2m+1} $ has a unique solution in the positive integers. Since $ m > 0 $, the right hand side is divisible by $ 4 $, therefore so is the left hand side. Looking at the congruence $ x^2 + y^2 \equiv 0 \pmod{4} $, we deduce that both $ x $ and $ y $ must be even. But then,

$$ \left(\frac{x}{2}\right)^2 + \left(\frac{y}{2}\right)^2 = 2^{2k+1} $$

and by the inductive hypothesis we know that the positive integer solutions to this equation are unique, given by $ x/2 = y/2 = 2^k $. Multiplying through by $ 2 $ then gives the desired result.

Now, apply this claim to your specific problem, with $ k = 2015 $.

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  • $\begingroup$ I don't know that concept so please tell me in an alternate method. $\endgroup$ – sai saandeep Sep 20 '16 at 16:02
  • $\begingroup$ I added another approach. $\endgroup$ – Starfall Sep 20 '16 at 16:22
  • $\begingroup$ There might not be a better method. However maybe we can work this into a more introductory approach. I'll try and see what I can do. $\endgroup$ – fleablood Sep 20 '16 at 16:23
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    $\begingroup$ That second method is very good! " x^2+y^2≡0(mod4), we deduce that both x and y must be even" This might not be universally known $(2k)^2 \equiv 0 \mod 4$ and $(2k + 1)^2 \equiv 1 \mod 4$. $x^2 \not \equiv 3 \mod 4$ so $x^2 + y^2 = 0,1,2 \mod 4$ if i)both are even ii) one odd and one even iii) both odd respectively. $\endgroup$ – fleablood Sep 20 '16 at 16:31
  • $\begingroup$ In the final step of the second method, you multiply through by $2$. This clearly gives a solution, but how does it prove uniqueness? $\endgroup$ – Théophile Sep 20 '16 at 19:55

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