Is the number a square?

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

#Algebra

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

It is easy to show that both $m$ and $n$ are even.Let $m=2r$ and $n=2s$

Hence, $2^{2r}=a^{2}-3^{2s}=(a-3^{s})(a+3^{s})$

Hence, $a-3^{s}=2^{i}$ ... $(1)$ and $a+3^{s}=2^{2r-i}$ ... $(2)$

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

Thus, $a-3^{s}=2$ and $a+3^{s}=2^{2r-1}$ .Hence, $3^{s}=2^{2r-2}-1$ ... $(3)$

Suppose, $s>1$ .Then $r≥3$ .But then the equation $(3)$ is impossible since when divided by $8$ , the left hand side $3^{s}$ leaves a remainder $1$ or $3$ while the right hand side would leave the remainder $7$ .Thus $s=1$ is the only possibility.When $s=1$ ,i.e, $n=2$ ,we have the solution $2^{4}+3^{2}=25$ .Thus $(m,n)=(4,2)$ is the only solution.

Souryajit Roy - 6 years, 6 months ago

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 - 6 years, 6 months ago

How to show that both $m$ and $n$ are even?

Kenny Lau - 6 years, 6 months ago

$a^{2}≡0 or 1(mod 3)$ and $2^{m}+3^{n}≡2^{m}(mod 3)$ .But $2^{m}$ is not congruent $0$ modulo $3$ .

So, $2^{m}≡1(mod 3)$ which implies $m$ is even.Hence, $3^{n}≡a^{2}≡0 or 1 (mod 4)$ .But $4$ does not divides $3^{n}$ .So, $3^{n}≡1(mod4)$ which implies $n$ is even.

Souryajit Roy - 6 years, 6 months ago

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

Guilherme Dela Corte - 6 years, 6 months ago

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

Joel Tan - 6 years, 6 months ago

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

Shashank Rammoorthy - 6 years, 6 months ago

I just subtracted eqn $1$ from eqn $2$ and got $2\times3^{s}=2^{2r-i}-2^{i}=2^{i}(2^{2r-2i}-1)$ .

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

So, $3^{s}=2^{1-1}(2^{2r-2.1}-1)=2^{2r-2}-1$ , which is eqn $3$ .Then the solution is very clear.

Souryajit Roy - 6 years, 6 months ago

@Souryajit Roy – Thanks.

Shashank Rammoorthy - 6 years, 6 months ago

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

Venkata Vineeth - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$