Proving that the sum and difference of two squares (not equal to zero) can't both be squares.












7














I have the following task:




Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



I tried to write down this:



Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!










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    7














    I have the following task:




    Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




    For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



    I tried to write down this:



    Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!










    share|cite|improve this question



























      7












      7








      7


      4





      I have the following task:




      Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




      For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



      I tried to write down this:



      Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!










      share|cite|improve this question















      I have the following task:




      Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.




      For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?



      I tried to write down this:



      Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!







      elementary-number-theory






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      share|cite|improve this question








      edited Apr 25 '15 at 17:34









      kate

      5281717




      5281717










      asked Apr 22 '15 at 11:13









      Atvin

      2,8451825




      2,8451825






















          2 Answers
          2






          active

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          7














          If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



          It is known that this is impossible. See, for example, here.






          share|cite|improve this answer

















          • 1




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19



















          3














          This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



          Fibonacci's Lost Theorem $ $ The area of an integral
          pythagorean triangle is not a perfect square.



          A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



          Naturally associated with every SAP is a "half square triangle",
          ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
          $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



          which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



          With these concepts in mind, the proof is very easy:
          If there exists a pythagorean triangle with square area then it
          may be primitivized and its area remains square. Let its primitive
          parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



          $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



          Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
          Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
          yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



          Remark $ $ This doubling construction is ancient - already in Euclid. It may be
          viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






          share|cite|improve this answer





















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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

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            active

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            7














            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer

















            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19
















            7














            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer

















            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19














            7












            7








            7






            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.






            share|cite|improve this answer












            If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.



            It is known that this is impossible. See, for example, here.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 22 '15 at 11:18









            mathlove

            91.7k881214




            91.7k881214








            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19














            • 1




              Very nice, thanks! :)
              – Atvin
              Apr 22 '15 at 11:19








            1




            1




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19




            Very nice, thanks! :)
            – Atvin
            Apr 22 '15 at 11:19











            3














            This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



            Fibonacci's Lost Theorem $ $ The area of an integral
            pythagorean triangle is not a perfect square.



            A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



            Naturally associated with every SAP is a "half square triangle",
            ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
            $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



            which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



            With these concepts in mind, the proof is very easy:
            If there exists a pythagorean triangle with square area then it
            may be primitivized and its area remains square. Let its primitive
            parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



            $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



            Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
            Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
            yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



            Remark $ $ This doubling construction is ancient - already in Euclid. It may be
            viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






            share|cite|improve this answer


























              3














              This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



              Fibonacci's Lost Theorem $ $ The area of an integral
              pythagorean triangle is not a perfect square.



              A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



              Naturally associated with every SAP is a "half square triangle",
              ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
              $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



              which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



              With these concepts in mind, the proof is very easy:
              If there exists a pythagorean triangle with square area then it
              may be primitivized and its area remains square. Let its primitive
              parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



              $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



              Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
              Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
              yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



              Remark $ $ This doubling construction is ancient - already in Euclid. It may be
              viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






              share|cite|improve this answer
























                3












                3








                3






                This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



                Fibonacci's Lost Theorem $ $ The area of an integral
                pythagorean triangle is not a perfect square.



                A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



                Naturally associated with every SAP is a "half square triangle",
                ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
                $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



                which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



                With these concepts in mind, the proof is very easy:
                If there exists a pythagorean triangle with square area then it
                may be primitivized and its area remains square. Let its primitive
                parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



                $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



                Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
                Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
                yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



                Remark $ $ This doubling construction is ancient - already in Euclid. It may be
                viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $






                share|cite|improve this answer












                This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.



                Fibonacci's Lost Theorem $ $ The area of an integral
                pythagorean triangle is not a perfect square.



                A square arithmetic progression (SAP) is an AP $rm x^2, y^2, z^2 $ with a square stepsize $rm, s^2,, $ viz. $$rm x^2 xrightarrow{Large s^2} y^2 xrightarrow{Large s^2} z^2$$



                Naturally associated with every SAP is a "half square triangle",
                ie. doubling $rm z^2 + x^2 $ produces a triangle of square area $rm s^2,, $ viz.
                $rm (z + x)^2 + (z - x)^2, = 2 (z^2 + x^2) = 4 y^2 $



                which indeed has $ $ area $rm, = (z + x) (z - x)/2 = (z^2 - x^2)/2 = s^2 $



                With these concepts in mind, the proof is very easy:
                If there exists a pythagorean triangle with square area then it
                may be primitivized and its area remains square. Let its primitive
                parametrization be $rm:(a,b):$ and let its area be $rm:c^2,:$ namely



                $$rm frac{1}2 leg_1 leg_2, = frac{1}2 (2:a:b) (a^2-b^2) = (a!-!b) a (a!+!b) b = c^2 $$



                Since $rm:a:$ and $rm:b:$ are coprime of opposite parity, $rm a!-!b, a, a!+!b, b $ are coprime factors of a square, thus all must be squares.
                Hence $rm a!-!b, a, a!+!b $ form a SAP; doubling its half square triangle
                yields a triangle with smaller square area $rm b < c^2, $ hence descent. $ $ QED



                Remark $ $ This doubling construction is ancient - already in Euclid. It may be
                viewed as a composition of quadratic forms $rm (z^2 + x^2) (1^2 + 1^2):. $







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 4 '15 at 16:38









                Bill Dubuque

                208k29190627




                208k29190627






























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