Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| leq 3$












6












$begingroup$


Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| leq 3$.



I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd.



But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.



Can someone help me with this please?










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  • $begingroup$
    Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
    $endgroup$
    – Frpzzd
    Dec 1 '18 at 23:42
















6












$begingroup$


Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| leq 3$.



I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd.



But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.



Can someone help me with this please?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
    $endgroup$
    – Frpzzd
    Dec 1 '18 at 23:42














6












6








6


3



$begingroup$


Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| leq 3$.



I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd.



But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.



Can someone help me with this please?










share|cite|improve this question











$endgroup$




Find all non-negative integers $a, b$ satisfying $|4a^2 - b^{b+1}| leq 3$.



I have been trying a simpler case $4a^2 - b^{b+1} = 0$. I found that when $b$ is odd then $b^{b+1}$ have to be in form $b^{b+1} = c^2$ where $c$ is integer. So we have $(2a + c)(2a - c)=0 implies 2a = c$ (otherwise $a$ would be negative). However, $c$ is odd because $b$ is also odd (odd number times odd number is odd). But this means that $a$ (since $2a = c$) isn't integer and that is contradiction. Therefore in this simpler case $b$ can't be odd.



But above applies only for specific simpler case. For example for $a = b = 1$ original inequality holds.



Can someone help me with this please?







number-theory inequality modular-arithmetic diophantine-equations






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edited Dec 1 '18 at 23:47









Frpzzd

22.6k840108




22.6k840108










asked Dec 1 '18 at 23:18









Snip3rSnip3r

869




869












  • $begingroup$
    Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
    $endgroup$
    – Frpzzd
    Dec 1 '18 at 23:42


















  • $begingroup$
    Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
    $endgroup$
    – Frpzzd
    Dec 1 '18 at 23:42
















$begingroup$
Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
$endgroup$
– Frpzzd
Dec 1 '18 at 23:42




$begingroup$
Okay, I have posted an answer. Yowza, that was a fun question - thanks for that. :D
$endgroup$
– Frpzzd
Dec 1 '18 at 23:42










1 Answer
1






active

oldest

votes


















3












$begingroup$

For the sake of simplicity, I will consider positive integers $a,b$.



First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4cdot 4^{4c^2}cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
$$(a,b)=(4^{2c^2}cdot c^{4c^2+1},4c^2)$$





Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}ne 1$.



Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.





Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.



Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.





Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
$$(a,b)=(1,1)$$
and no others.



Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.





We are done! We have only the solutions $(1,1)$ and
$$(4^{2c^2}cdot c^{4c^2+1},4c^2)$$
for $cinmathbb N$.






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    $begingroup$

    For the sake of simplicity, I will consider positive integers $a,b$.



    First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4cdot 4^{4c^2}cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
    $$(a,b)=(4^{2c^2}cdot c^{4c^2+1},4c^2)$$





    Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}ne 1$.



    Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.





    Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.



    Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.





    Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
    $$(a,b)=(1,1)$$
    and no others.



    Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.





    We are done! We have only the solutions $(1,1)$ and
    $$(4^{2c^2}cdot c^{4c^2+1},4c^2)$$
    for $cinmathbb N$.






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      For the sake of simplicity, I will consider positive integers $a,b$.



      First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4cdot 4^{4c^2}cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
      $$(a,b)=(4^{2c^2}cdot c^{4c^2+1},4c^2)$$





      Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}ne 1$.



      Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.





      Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.



      Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.





      Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
      $$(a,b)=(1,1)$$
      and no others.



      Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.





      We are done! We have only the solutions $(1,1)$ and
      $$(4^{2c^2}cdot c^{4c^2+1},4c^2)$$
      for $cinmathbb N$.






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        For the sake of simplicity, I will consider positive integers $a,b$.



        First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4cdot 4^{4c^2}cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
        $$(a,b)=(4^{2c^2}cdot c^{4c^2+1},4c^2)$$





        Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}ne 1$.



        Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.





        Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.



        Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.





        Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
        $$(a,b)=(1,1)$$
        and no others.



        Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.





        We are done! We have only the solutions $(1,1)$ and
        $$(4^{2c^2}cdot c^{4c^2+1},4c^2)$$
        for $cinmathbb N$.






        share|cite|improve this answer









        $endgroup$



        For the sake of simplicity, I will consider positive integers $a,b$.



        First, let's consider the case of $4a^2-b^{b+1}=0$. You've already noticed that $b$ cannot be odd, and so we shall consider only the case of even $b$. If $b$ is even, then $b+1$ is odd and so $b^{b+1}$ is a perfect square iff $b$ is a perfect square, so we have that $b$ is a perfect square. Since it is an even perfect square, we may write $b=4c^2$ and $b^{b+1}=4cdot 4^{4c^2}cdot c^{8c^2+2}$ and we may let $a=4^{2c^2}cdot c^{4c^2+1}$ for a solution. Thus, we have found one possible solution set:
        $$(a,b)=(4^{2c^2}cdot c^{4c^2+1},4c^2)$$





        Now suppose that $4a^2-b^{b+1}=1$. Then $(2a+1)(2a-1)=b^{b+1}$, and so $b$ must be odd, meaning that $b+1$ is even and $b^{b+1}$ is a perfect square. But $4a^2=b^{b+1}+1$ is also a perfect square, giving a contradiction. So $4a^2-b^{b+1}ne 1$.



        Similarly, suppose that $4a^2-b^{b+1}=-1$, or $4a^2=b^{b+1}-1$. We have again that $b$ must be odd, and so $b+1$ is even and $b^{b+1}$ is a perfect square, which is a contradiction since $b^{b+1}-1$ is a perfect square.





        Now suppose that $4a^2-b^{b+1}=2$, or $4a^2=b^{b+1}+2$. Then $b$ must be even, and we have that the $2$-adic valuation of $b^{b+1}+2$ is equal to $1$. But the $2$-adic valuation of $4a^2$ is at least $2$, giving a contradiction.



        Use the same reasoning for the case of $4a^2-b^{b+1}=-2$.





        Suppose that $4a^2-b^{b+1}=3$, or $4a^2=b^{b+1}+3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}+3$ is also a perfect square, which is only possible if $b^{b+1}=1$. This gives the solution pair
        $$(a,b)=(1,1)$$
        and no others.



        Finally, suppose that $4a^2-b^{b+1}=-3$ or $4a^2=b^{b+1}-3$. Then $b$ must be odd and $b+1$ must be even, so $b^{b+1}$ is a perfect square. However, $b^{b+1}-3$ is also a perfect square, which can only occur if $b^{b+1}=4$, but this never happens for positive integers $b$.





        We are done! We have only the solutions $(1,1)$ and
        $$(4^{2c^2}cdot c^{4c^2+1},4c^2)$$
        for $cinmathbb N$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 23:39









        FrpzzdFrpzzd

        22.6k840108




        22.6k840108






























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