On $xy$ vs $ab$ where $2xy+x+y+1 = 2ab+a+b$












0














Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.



Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.



I was wondering what we can say about the relation between $x*y$ and $a*b$.
As $n$ and $n+1$ are very close, and as $xy > x+y$, there is likely to be a maximal difference between $xy$ and $ab$.



Am I right to suppose that to fulfil the equations, $xy = ab$ or $xy + 1 = ab$ ? So this maximum difference is 1 ?










share|cite|improve this question



























    0














    Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.



    Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.



    I was wondering what we can say about the relation between $x*y$ and $a*b$.
    As $n$ and $n+1$ are very close, and as $xy > x+y$, there is likely to be a maximal difference between $xy$ and $ab$.



    Am I right to suppose that to fulfil the equations, $xy = ab$ or $xy + 1 = ab$ ? So this maximum difference is 1 ?










    share|cite|improve this question

























      0












      0








      0







      Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.



      Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.



      I was wondering what we can say about the relation between $x*y$ and $a*b$.
      As $n$ and $n+1$ are very close, and as $xy > x+y$, there is likely to be a maximal difference between $xy$ and $ab$.



      Am I right to suppose that to fulfil the equations, $xy = ab$ or $xy + 1 = ab$ ? So this maximum difference is 1 ?










      share|cite|improve this question













      Let us suppose that $2xy+x+y = n$, where $x,y,n$ are positive integers and $x≥2$ and $y≥2$.



      Let us choose x and y, so that $n+1 = 2ab+a+b$ , where $a, b$ are also positive integer numbers.



      I was wondering what we can say about the relation between $x*y$ and $a*b$.
      As $n$ and $n+1$ are very close, and as $xy > x+y$, there is likely to be a maximal difference between $xy$ and $ab$.



      Am I right to suppose that to fulfil the equations, $xy = ab$ or $xy + 1 = ab$ ? So this maximum difference is 1 ?







      elementary-number-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 20:07









      Tilsight

      135




      135






















          3 Answers
          3






          active

          oldest

          votes


















          0














          apparently not



           a: 4 b: 1 x: 2 y: 2 |a*b - x*y| : 0
          a: 5 b: 3 x: 7 y: 2 |a*b - x*y| : 1
          a: 5 b: 5 x: 8 y: 3 |a*b - x*y| : 1
          a: 6 b: 2 x: 4 y: 3 |a*b - x*y| : 0
          a: 6 b: 4 x: 11 y: 2 |a*b - x*y| : 2
          a: 7 b: 4 x: 9 y: 3 |a*b - x*y| : 1
          a: 8 b: 1 x: 3 y: 3 |a*b - x*y| : 1
          a: 8 b: 3 x: 6 y: 4 |a*b - x*y| : 0
          a: 8 b: 5 x: 18 y: 2 |a*b - x*y| : 4
          a: 8 b: 7 x: 11 y: 5 |a*b - x*y| : 1
          a: 8 b: 8 x: 20 y: 3 |a*b - x*y| : 4
          a: 9 b: 1 x: 5 y: 2 |a*b - x*y| : 1
          a: 9 b: 4 x: 6 y: 6 |a*b - x*y| : 0
          a: 9 b: 5 x: 11 y: 4 |a*b - x*y| : 1
          a: 9 b: 6 x: 24 y: 2 |a*b - x*y| : 6
          a: 9 b: 6 x: 17 y: 3 |a*b - x*y| : 3
          a: 10 b: 3 x: 14 y: 2 |a*b - x*y| : 2
          a: 10 b: 4 x: 8 y: 5 |a*b - x*y| : 0
          a: 10 b: 8 x: 35 y: 2 |a*b - x*y| : 10





          share|cite|improve this answer





















          • Thank you, Will.
            – Tilsight
            Nov 25 at 20:41










          • Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
            – Tilsight
            Nov 25 at 20:43





















          0














          Let for instance $a=1$ so we get that $3 b =n$ , and assume that $x =y$ so we get that $2 x^2 +2 x =n = 3b $ which leads to $x = frac{-2+sqrt{4+24b}}{4}$,



          so for instance : for $b=28$ gives that $x = y = 6$ and $a=1$, so $| x y -a b| = 8$ , and you can get the maximum difference as large as you want.






          share|cite|improve this answer





























            0














            Not bounded at all.



            With $a equiv 3,4 pmod 5,$ we can take $b = a-3,$ then $y=2$ and
            $$ x = frac{2(a-3)(a+1)}{5} $$



            With $a equiv 0,1 pmod 5,$ we can take $b = a-2,$ then $y=2$ and
            $$ x = frac{2a^2 -2a-5}{5} $$



             a: 275 b: 273 x: 30139 y: 2 |a*b - x*y| : 14797
            a: 276 b: 274 x: 30359 y: 2 |a*b - x*y| : 14906
            a: 278 b: 275 x: 30690 y: 2 |a*b - x*y| : 15070
            a: 279 b: 276 x: 30912 y: 2 |a*b - x*y| : 15180
            a: 280 b: 278 x: 31247 y: 2 |a*b - x*y| : 15346
            a: 281 b: 279 x: 31471 y: 2 |a*b - x*y| : 15457
            a: 283 b: 280 x: 31808 y: 2 |a*b - x*y| : 15624
            a: 284 b: 281 x: 32034 y: 2 |a*b - x*y| : 15736
            a: 285 b: 283 x: 32375 y: 2 |a*b - x*y| : 15905
            a: 286 b: 284 x: 32603 y: 2 |a*b - x*y| : 16018
            a: 288 b: 285 x: 32946 y: 2 |a*b - x*y| : 16188
            a: 289 b: 286 x: 33176 y: 2 |a*b - x*y| : 16302
            a: 290 b: 288 x: 33523 y: 2 |a*b - x*y| : 16474
            a: 291 b: 289 x: 33755 y: 2 |a*b - x*y| : 16589
            a: 293 b: 290 x: 34104 y: 2 |a*b - x*y| : 16762
            a: 294 b: 291 x: 34338 y: 2 |a*b - x*y| : 16878
            a: 295 b: 293 x: 34691 y: 2 |a*b - x*y| : 17053
            a: 296 b: 294 x: 34927 y: 2 |a*b - x*y| : 17170
            a: 298 b: 295 x: 35282 y: 2 |a*b - x*y| : 17346
            a: 299 b: 296 x: 35520 y: 2 |a*b - x*y| : 17464
            a: 300 b: 298 x: 35879 y: 2 |a*b - x*y| : 17642





            share|cite|improve this answer





















            • Thanks guys. Respect 😀
              – Tilsight
              Nov 25 at 20:57











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013321%2fon-xy-vs-ab-where-2xyxy1-2abab%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            apparently not



             a: 4 b: 1 x: 2 y: 2 |a*b - x*y| : 0
            a: 5 b: 3 x: 7 y: 2 |a*b - x*y| : 1
            a: 5 b: 5 x: 8 y: 3 |a*b - x*y| : 1
            a: 6 b: 2 x: 4 y: 3 |a*b - x*y| : 0
            a: 6 b: 4 x: 11 y: 2 |a*b - x*y| : 2
            a: 7 b: 4 x: 9 y: 3 |a*b - x*y| : 1
            a: 8 b: 1 x: 3 y: 3 |a*b - x*y| : 1
            a: 8 b: 3 x: 6 y: 4 |a*b - x*y| : 0
            a: 8 b: 5 x: 18 y: 2 |a*b - x*y| : 4
            a: 8 b: 7 x: 11 y: 5 |a*b - x*y| : 1
            a: 8 b: 8 x: 20 y: 3 |a*b - x*y| : 4
            a: 9 b: 1 x: 5 y: 2 |a*b - x*y| : 1
            a: 9 b: 4 x: 6 y: 6 |a*b - x*y| : 0
            a: 9 b: 5 x: 11 y: 4 |a*b - x*y| : 1
            a: 9 b: 6 x: 24 y: 2 |a*b - x*y| : 6
            a: 9 b: 6 x: 17 y: 3 |a*b - x*y| : 3
            a: 10 b: 3 x: 14 y: 2 |a*b - x*y| : 2
            a: 10 b: 4 x: 8 y: 5 |a*b - x*y| : 0
            a: 10 b: 8 x: 35 y: 2 |a*b - x*y| : 10





            share|cite|improve this answer





















            • Thank you, Will.
              – Tilsight
              Nov 25 at 20:41










            • Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
              – Tilsight
              Nov 25 at 20:43


















            0














            apparently not



             a: 4 b: 1 x: 2 y: 2 |a*b - x*y| : 0
            a: 5 b: 3 x: 7 y: 2 |a*b - x*y| : 1
            a: 5 b: 5 x: 8 y: 3 |a*b - x*y| : 1
            a: 6 b: 2 x: 4 y: 3 |a*b - x*y| : 0
            a: 6 b: 4 x: 11 y: 2 |a*b - x*y| : 2
            a: 7 b: 4 x: 9 y: 3 |a*b - x*y| : 1
            a: 8 b: 1 x: 3 y: 3 |a*b - x*y| : 1
            a: 8 b: 3 x: 6 y: 4 |a*b - x*y| : 0
            a: 8 b: 5 x: 18 y: 2 |a*b - x*y| : 4
            a: 8 b: 7 x: 11 y: 5 |a*b - x*y| : 1
            a: 8 b: 8 x: 20 y: 3 |a*b - x*y| : 4
            a: 9 b: 1 x: 5 y: 2 |a*b - x*y| : 1
            a: 9 b: 4 x: 6 y: 6 |a*b - x*y| : 0
            a: 9 b: 5 x: 11 y: 4 |a*b - x*y| : 1
            a: 9 b: 6 x: 24 y: 2 |a*b - x*y| : 6
            a: 9 b: 6 x: 17 y: 3 |a*b - x*y| : 3
            a: 10 b: 3 x: 14 y: 2 |a*b - x*y| : 2
            a: 10 b: 4 x: 8 y: 5 |a*b - x*y| : 0
            a: 10 b: 8 x: 35 y: 2 |a*b - x*y| : 10





            share|cite|improve this answer





















            • Thank you, Will.
              – Tilsight
              Nov 25 at 20:41










            • Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
              – Tilsight
              Nov 25 at 20:43
















            0












            0








            0






            apparently not



             a: 4 b: 1 x: 2 y: 2 |a*b - x*y| : 0
            a: 5 b: 3 x: 7 y: 2 |a*b - x*y| : 1
            a: 5 b: 5 x: 8 y: 3 |a*b - x*y| : 1
            a: 6 b: 2 x: 4 y: 3 |a*b - x*y| : 0
            a: 6 b: 4 x: 11 y: 2 |a*b - x*y| : 2
            a: 7 b: 4 x: 9 y: 3 |a*b - x*y| : 1
            a: 8 b: 1 x: 3 y: 3 |a*b - x*y| : 1
            a: 8 b: 3 x: 6 y: 4 |a*b - x*y| : 0
            a: 8 b: 5 x: 18 y: 2 |a*b - x*y| : 4
            a: 8 b: 7 x: 11 y: 5 |a*b - x*y| : 1
            a: 8 b: 8 x: 20 y: 3 |a*b - x*y| : 4
            a: 9 b: 1 x: 5 y: 2 |a*b - x*y| : 1
            a: 9 b: 4 x: 6 y: 6 |a*b - x*y| : 0
            a: 9 b: 5 x: 11 y: 4 |a*b - x*y| : 1
            a: 9 b: 6 x: 24 y: 2 |a*b - x*y| : 6
            a: 9 b: 6 x: 17 y: 3 |a*b - x*y| : 3
            a: 10 b: 3 x: 14 y: 2 |a*b - x*y| : 2
            a: 10 b: 4 x: 8 y: 5 |a*b - x*y| : 0
            a: 10 b: 8 x: 35 y: 2 |a*b - x*y| : 10





            share|cite|improve this answer












            apparently not



             a: 4 b: 1 x: 2 y: 2 |a*b - x*y| : 0
            a: 5 b: 3 x: 7 y: 2 |a*b - x*y| : 1
            a: 5 b: 5 x: 8 y: 3 |a*b - x*y| : 1
            a: 6 b: 2 x: 4 y: 3 |a*b - x*y| : 0
            a: 6 b: 4 x: 11 y: 2 |a*b - x*y| : 2
            a: 7 b: 4 x: 9 y: 3 |a*b - x*y| : 1
            a: 8 b: 1 x: 3 y: 3 |a*b - x*y| : 1
            a: 8 b: 3 x: 6 y: 4 |a*b - x*y| : 0
            a: 8 b: 5 x: 18 y: 2 |a*b - x*y| : 4
            a: 8 b: 7 x: 11 y: 5 |a*b - x*y| : 1
            a: 8 b: 8 x: 20 y: 3 |a*b - x*y| : 4
            a: 9 b: 1 x: 5 y: 2 |a*b - x*y| : 1
            a: 9 b: 4 x: 6 y: 6 |a*b - x*y| : 0
            a: 9 b: 5 x: 11 y: 4 |a*b - x*y| : 1
            a: 9 b: 6 x: 24 y: 2 |a*b - x*y| : 6
            a: 9 b: 6 x: 17 y: 3 |a*b - x*y| : 3
            a: 10 b: 3 x: 14 y: 2 |a*b - x*y| : 2
            a: 10 b: 4 x: 8 y: 5 |a*b - x*y| : 0
            a: 10 b: 8 x: 35 y: 2 |a*b - x*y| : 10






            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 25 at 20:24









            Will Jagy

            101k598198




            101k598198












            • Thank you, Will.
              – Tilsight
              Nov 25 at 20:41










            • Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
              – Tilsight
              Nov 25 at 20:43




















            • Thank you, Will.
              – Tilsight
              Nov 25 at 20:41










            • Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
              – Tilsight
              Nov 25 at 20:43


















            Thank you, Will.
            – Tilsight
            Nov 25 at 20:41




            Thank you, Will.
            – Tilsight
            Nov 25 at 20:41












            Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
            – Tilsight
            Nov 25 at 20:43






            Just a question, you gave a very quick answer with a lot of examples. How did you do it so effectively? U wrote a script in C?
            – Tilsight
            Nov 25 at 20:43













            0














            Let for instance $a=1$ so we get that $3 b =n$ , and assume that $x =y$ so we get that $2 x^2 +2 x =n = 3b $ which leads to $x = frac{-2+sqrt{4+24b}}{4}$,



            so for instance : for $b=28$ gives that $x = y = 6$ and $a=1$, so $| x y -a b| = 8$ , and you can get the maximum difference as large as you want.






            share|cite|improve this answer


























              0














              Let for instance $a=1$ so we get that $3 b =n$ , and assume that $x =y$ so we get that $2 x^2 +2 x =n = 3b $ which leads to $x = frac{-2+sqrt{4+24b}}{4}$,



              so for instance : for $b=28$ gives that $x = y = 6$ and $a=1$, so $| x y -a b| = 8$ , and you can get the maximum difference as large as you want.






              share|cite|improve this answer
























                0












                0








                0






                Let for instance $a=1$ so we get that $3 b =n$ , and assume that $x =y$ so we get that $2 x^2 +2 x =n = 3b $ which leads to $x = frac{-2+sqrt{4+24b}}{4}$,



                so for instance : for $b=28$ gives that $x = y = 6$ and $a=1$, so $| x y -a b| = 8$ , and you can get the maximum difference as large as you want.






                share|cite|improve this answer












                Let for instance $a=1$ so we get that $3 b =n$ , and assume that $x =y$ so we get that $2 x^2 +2 x =n = 3b $ which leads to $x = frac{-2+sqrt{4+24b}}{4}$,



                so for instance : for $b=28$ gives that $x = y = 6$ and $a=1$, so $| x y -a b| = 8$ , and you can get the maximum difference as large as you want.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 at 20:42









                Ahmad

                2,5241625




                2,5241625























                    0














                    Not bounded at all.



                    With $a equiv 3,4 pmod 5,$ we can take $b = a-3,$ then $y=2$ and
                    $$ x = frac{2(a-3)(a+1)}{5} $$



                    With $a equiv 0,1 pmod 5,$ we can take $b = a-2,$ then $y=2$ and
                    $$ x = frac{2a^2 -2a-5}{5} $$



                     a: 275 b: 273 x: 30139 y: 2 |a*b - x*y| : 14797
                    a: 276 b: 274 x: 30359 y: 2 |a*b - x*y| : 14906
                    a: 278 b: 275 x: 30690 y: 2 |a*b - x*y| : 15070
                    a: 279 b: 276 x: 30912 y: 2 |a*b - x*y| : 15180
                    a: 280 b: 278 x: 31247 y: 2 |a*b - x*y| : 15346
                    a: 281 b: 279 x: 31471 y: 2 |a*b - x*y| : 15457
                    a: 283 b: 280 x: 31808 y: 2 |a*b - x*y| : 15624
                    a: 284 b: 281 x: 32034 y: 2 |a*b - x*y| : 15736
                    a: 285 b: 283 x: 32375 y: 2 |a*b - x*y| : 15905
                    a: 286 b: 284 x: 32603 y: 2 |a*b - x*y| : 16018
                    a: 288 b: 285 x: 32946 y: 2 |a*b - x*y| : 16188
                    a: 289 b: 286 x: 33176 y: 2 |a*b - x*y| : 16302
                    a: 290 b: 288 x: 33523 y: 2 |a*b - x*y| : 16474
                    a: 291 b: 289 x: 33755 y: 2 |a*b - x*y| : 16589
                    a: 293 b: 290 x: 34104 y: 2 |a*b - x*y| : 16762
                    a: 294 b: 291 x: 34338 y: 2 |a*b - x*y| : 16878
                    a: 295 b: 293 x: 34691 y: 2 |a*b - x*y| : 17053
                    a: 296 b: 294 x: 34927 y: 2 |a*b - x*y| : 17170
                    a: 298 b: 295 x: 35282 y: 2 |a*b - x*y| : 17346
                    a: 299 b: 296 x: 35520 y: 2 |a*b - x*y| : 17464
                    a: 300 b: 298 x: 35879 y: 2 |a*b - x*y| : 17642





                    share|cite|improve this answer





















                    • Thanks guys. Respect 😀
                      – Tilsight
                      Nov 25 at 20:57
















                    0














                    Not bounded at all.



                    With $a equiv 3,4 pmod 5,$ we can take $b = a-3,$ then $y=2$ and
                    $$ x = frac{2(a-3)(a+1)}{5} $$



                    With $a equiv 0,1 pmod 5,$ we can take $b = a-2,$ then $y=2$ and
                    $$ x = frac{2a^2 -2a-5}{5} $$



                     a: 275 b: 273 x: 30139 y: 2 |a*b - x*y| : 14797
                    a: 276 b: 274 x: 30359 y: 2 |a*b - x*y| : 14906
                    a: 278 b: 275 x: 30690 y: 2 |a*b - x*y| : 15070
                    a: 279 b: 276 x: 30912 y: 2 |a*b - x*y| : 15180
                    a: 280 b: 278 x: 31247 y: 2 |a*b - x*y| : 15346
                    a: 281 b: 279 x: 31471 y: 2 |a*b - x*y| : 15457
                    a: 283 b: 280 x: 31808 y: 2 |a*b - x*y| : 15624
                    a: 284 b: 281 x: 32034 y: 2 |a*b - x*y| : 15736
                    a: 285 b: 283 x: 32375 y: 2 |a*b - x*y| : 15905
                    a: 286 b: 284 x: 32603 y: 2 |a*b - x*y| : 16018
                    a: 288 b: 285 x: 32946 y: 2 |a*b - x*y| : 16188
                    a: 289 b: 286 x: 33176 y: 2 |a*b - x*y| : 16302
                    a: 290 b: 288 x: 33523 y: 2 |a*b - x*y| : 16474
                    a: 291 b: 289 x: 33755 y: 2 |a*b - x*y| : 16589
                    a: 293 b: 290 x: 34104 y: 2 |a*b - x*y| : 16762
                    a: 294 b: 291 x: 34338 y: 2 |a*b - x*y| : 16878
                    a: 295 b: 293 x: 34691 y: 2 |a*b - x*y| : 17053
                    a: 296 b: 294 x: 34927 y: 2 |a*b - x*y| : 17170
                    a: 298 b: 295 x: 35282 y: 2 |a*b - x*y| : 17346
                    a: 299 b: 296 x: 35520 y: 2 |a*b - x*y| : 17464
                    a: 300 b: 298 x: 35879 y: 2 |a*b - x*y| : 17642





                    share|cite|improve this answer





















                    • Thanks guys. Respect 😀
                      – Tilsight
                      Nov 25 at 20:57














                    0












                    0








                    0






                    Not bounded at all.



                    With $a equiv 3,4 pmod 5,$ we can take $b = a-3,$ then $y=2$ and
                    $$ x = frac{2(a-3)(a+1)}{5} $$



                    With $a equiv 0,1 pmod 5,$ we can take $b = a-2,$ then $y=2$ and
                    $$ x = frac{2a^2 -2a-5}{5} $$



                     a: 275 b: 273 x: 30139 y: 2 |a*b - x*y| : 14797
                    a: 276 b: 274 x: 30359 y: 2 |a*b - x*y| : 14906
                    a: 278 b: 275 x: 30690 y: 2 |a*b - x*y| : 15070
                    a: 279 b: 276 x: 30912 y: 2 |a*b - x*y| : 15180
                    a: 280 b: 278 x: 31247 y: 2 |a*b - x*y| : 15346
                    a: 281 b: 279 x: 31471 y: 2 |a*b - x*y| : 15457
                    a: 283 b: 280 x: 31808 y: 2 |a*b - x*y| : 15624
                    a: 284 b: 281 x: 32034 y: 2 |a*b - x*y| : 15736
                    a: 285 b: 283 x: 32375 y: 2 |a*b - x*y| : 15905
                    a: 286 b: 284 x: 32603 y: 2 |a*b - x*y| : 16018
                    a: 288 b: 285 x: 32946 y: 2 |a*b - x*y| : 16188
                    a: 289 b: 286 x: 33176 y: 2 |a*b - x*y| : 16302
                    a: 290 b: 288 x: 33523 y: 2 |a*b - x*y| : 16474
                    a: 291 b: 289 x: 33755 y: 2 |a*b - x*y| : 16589
                    a: 293 b: 290 x: 34104 y: 2 |a*b - x*y| : 16762
                    a: 294 b: 291 x: 34338 y: 2 |a*b - x*y| : 16878
                    a: 295 b: 293 x: 34691 y: 2 |a*b - x*y| : 17053
                    a: 296 b: 294 x: 34927 y: 2 |a*b - x*y| : 17170
                    a: 298 b: 295 x: 35282 y: 2 |a*b - x*y| : 17346
                    a: 299 b: 296 x: 35520 y: 2 |a*b - x*y| : 17464
                    a: 300 b: 298 x: 35879 y: 2 |a*b - x*y| : 17642





                    share|cite|improve this answer












                    Not bounded at all.



                    With $a equiv 3,4 pmod 5,$ we can take $b = a-3,$ then $y=2$ and
                    $$ x = frac{2(a-3)(a+1)}{5} $$



                    With $a equiv 0,1 pmod 5,$ we can take $b = a-2,$ then $y=2$ and
                    $$ x = frac{2a^2 -2a-5}{5} $$



                     a: 275 b: 273 x: 30139 y: 2 |a*b - x*y| : 14797
                    a: 276 b: 274 x: 30359 y: 2 |a*b - x*y| : 14906
                    a: 278 b: 275 x: 30690 y: 2 |a*b - x*y| : 15070
                    a: 279 b: 276 x: 30912 y: 2 |a*b - x*y| : 15180
                    a: 280 b: 278 x: 31247 y: 2 |a*b - x*y| : 15346
                    a: 281 b: 279 x: 31471 y: 2 |a*b - x*y| : 15457
                    a: 283 b: 280 x: 31808 y: 2 |a*b - x*y| : 15624
                    a: 284 b: 281 x: 32034 y: 2 |a*b - x*y| : 15736
                    a: 285 b: 283 x: 32375 y: 2 |a*b - x*y| : 15905
                    a: 286 b: 284 x: 32603 y: 2 |a*b - x*y| : 16018
                    a: 288 b: 285 x: 32946 y: 2 |a*b - x*y| : 16188
                    a: 289 b: 286 x: 33176 y: 2 |a*b - x*y| : 16302
                    a: 290 b: 288 x: 33523 y: 2 |a*b - x*y| : 16474
                    a: 291 b: 289 x: 33755 y: 2 |a*b - x*y| : 16589
                    a: 293 b: 290 x: 34104 y: 2 |a*b - x*y| : 16762
                    a: 294 b: 291 x: 34338 y: 2 |a*b - x*y| : 16878
                    a: 295 b: 293 x: 34691 y: 2 |a*b - x*y| : 17053
                    a: 296 b: 294 x: 34927 y: 2 |a*b - x*y| : 17170
                    a: 298 b: 295 x: 35282 y: 2 |a*b - x*y| : 17346
                    a: 299 b: 296 x: 35520 y: 2 |a*b - x*y| : 17464
                    a: 300 b: 298 x: 35879 y: 2 |a*b - x*y| : 17642






                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 25 at 20:44









                    Will Jagy

                    101k598198




                    101k598198












                    • Thanks guys. Respect 😀
                      – Tilsight
                      Nov 25 at 20:57


















                    • Thanks guys. Respect 😀
                      – Tilsight
                      Nov 25 at 20:57
















                    Thanks guys. Respect 😀
                    – Tilsight
                    Nov 25 at 20:57




                    Thanks guys. Respect 😀
                    – Tilsight
                    Nov 25 at 20:57


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013321%2fon-xy-vs-ab-where-2xyxy1-2abab%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Bundesstraße 106

                    Verónica Boquete

                    Ida-Boy-Ed-Garten