Is the number a square?

Find all pairs (m,n) of positive integers for which the above expression is a perfect square.

Note by Shashank Rammoorthy
4 years, 11 months ago

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Suppose 2m+3n=a22^{m}+3^{n}=a^{2}.

It is easy to show that both mm and nn are even.Let m=2rm=2r and n=2sn=2s

Hence, 22r=a232s=(a3s)(a+3s)2^{2r}=a^{2}-3^{2s}=(a-3^{s})(a+3^{s})

Hence, a3s=2ia-3^{s}=2^{i}...(1)(1) and a+3s=22ria+3^{s}=2^{2r-i}...(2)(2)

(2)(1)(2)-(1) gives 2.3s=2i(22r2i1)2.3^{s}=2^{i}(2^{2r-2i}-1), which implies i=1i=1.

Thus, a3s=2a-3^{s}=2 and a+3s=22r1a+3^{s}=2^{2r-1}.Hence, 3s=22r213^{s}=2^{2r-2}-1...(3)(3)

Suppose, s>1s>1.Then r3r≥3.But then the equation (3)(3) is impossible since when divided by 88, the left hand side 3s3^{s} leaves a remainder 11 or 33 while the right hand side would leave the remainder 77.Thus s=1s=1 is the only possibility.When s=1s=1,i.e, n=2n=2,we have the solution 24+32=252^{4}+3^{2}=25.Thus (m,n)=(4,2)(m,n)=(4,2) is the only solution.

Souryajit Roy - 4 years, 11 months ago

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Thank you for this solution! Been thinking about this all day. Never got the chance to sit down with a pen and paper, unfortunately...

Ryan Tamburrino - 4 years, 11 months ago

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How to show that both mm and nn are even?

Kenny Lau - 4 years, 11 months ago

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a20or1(mod3)a^{2}≡0 or 1(mod 3) and 2m+3n2m(mod3)2^{m}+3^{n}≡2^{m}(mod 3).But 2m2^{m} is not congruent 00 modulo 33.

So,2m1(mod3)2^{m}≡1(mod 3) which implies mm is even.Hence, 3na20or1(mod4)3^{n}≡a^{2}≡0 or 1 (mod 4).But 44 does not divides 3n3^{n}.So,3n1(mod4)3^{n}≡1(mod4) which implies nn is even.

Souryajit Roy - 4 years, 11 months ago

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How about (3,0) (3,0) and (0,1) (0,1) ?

Guilherme Dela Corte - 4 years, 11 months ago

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m, n are positive integers. 0 is not positive.

Joel Tan - 4 years, 11 months ago

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Could you please explain your solution from line 5 onwards? How is it 2.3^s? Thanks.

Shashank Rammoorthy - 4 years, 11 months ago

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I just subtracted eqn 11 from eqn 22 and got 2×3s=22ri2i=2i(22r2i1)2\times3^{s}=2^{2r-i}-2^{i}=2^{i}(2^{2r-2i}-1).

Hence, 3s=2i1(22r2i1)3^{s}=2^{i-1}(2^{2r-2i}-1).If i>1i>1,then 22 divides 3s3^{s}, which is impossible.So i=1i=1.

So,3s=211(22r2.11)=22r213^{s}=2^{1-1}(2^{2r-2.1}-1)=2^{2r-2}-1, which is eqn 33.Then the solution is very clear.

Souryajit Roy - 4 years, 11 months ago

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@Souryajit Roy Thanks.

Shashank Rammoorthy - 4 years, 11 months ago

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(4,2),(3,0),(0,1),

Venkata Vineeth - 4 years, 11 months ago

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