How to solve this

Determine all integers n which satisfy

$((25/2+((625/4)-n)^\frac{1}{2})^\frac{1}{2})+((25/2-((625/4)-n)^\frac{1}{2})^\frac{1}{2})$

is an integer

#Algebra #MathProblem #Math

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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Comments

If $b \; = \; \sqrt{a + \sqrt{a^2-n}} + \sqrt{a - \sqrt{a^2-n}}$ then $b^2 \; = \; a + \sqrt{a^2-n} + 2\sqrt{n} + a - \sqrt{a^2-n} \; = \; 2a + 2\sqrt{n}$ so that $b^2 \ge 2a = 25$ and $n \; = \; \big(\tfrac12b^2 - a\big)^2 \; = \; \big(\tfrac{b^2-25}{2}\big)^2$ Thus we need $b$ to be an odd integer greater than or equal to $5$ . and so $n \; = \; \big(\tfrac{(2m+3)^2-25}{2}\big)^2 \; = \; (2m^2 + 6m - 8)^2 \; = \; 4(m-1)^2(m+4)^2$ for any $m \ge 1$ .

Mark Hennings - 7 years, 7 months ago

square the expression to get $25+\sqrt{n}$ , so we try to find all n such that that expression is a perfect square. this gives $11^2, 24^2, 35^2, ..., (m^2-25)^2$ for all int m > 4

Liam Lawson - 7 years, 7 months ago

n is 144 !!! nt for any n > 4 ; say as 5 will nt yield an integer .. liam reconsider your solution.. !!

Ramesh Goenka - 7 years, 7 months ago

even n=0 wrks gud.. !!

Ramesh Goenka - 7 years, 7 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}`	$\boxed{123}$