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# 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

Note by Suci Mayeza
4 years, 2 months ago

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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 $$m \ge 1$$.

- 4 years, 2 months ago

even n=0 wrks gud.. !!

- 4 years, 2 months ago

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

- 4 years, 2 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

- 4 years, 2 months ago