MHB Age Problem Need Detailed Solutions

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The age of Mark now times the age of Amy a year from now is the square of an
integer. The age of Mark a year from now times the age of Amy now is also the
square of an integer. If Amy is 8 years old now, and Mark is now older than 1 but
younger than 100, how old is Mark now?
 
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Marcelo Arevalo said:
The age of Mark now times the age of Amy a year from now is the square of an
integer. The age of Mark a year from now times the age of Amy now is also the
square of an integer. If Amy is 8 years old now, and Mark is now older than 1 but
younger than 100, how old is Mark now?

Write out what you know. Let x be the age of mark and y the age of Amy.

Then x( y + 1) = what? Then what is your other equation?
 
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Let : X = Mark age 2 to 99
Y = Amy age = 2

X (Y+1) = \sqrt{A} EQ 1
(X+1) Y = \sqrt{B} EQ 2

Substituting:
X (8+1) = \sqrt{A}
(X+1) 8 = \sqrt{B}

now I am stucked..its a dead end for me.
 
Marcelo Arevalo said:
Let : X = Mark age 2 to 99
Y = Amy age = 2

X (Y+1) = \sqrt{A} EQ 1
(X+1) Y = \sqrt{B} EQ 2

Substituting:
X (8+1) = \sqrt{A}
(X+1) 8 = \sqrt{B}

now I am stucked..its a dead end for me.

Then we have
\begin{align}
x(y+1) &= A\\
x(x + 8 + 1) &= A\\
x^2 + 9x &= A\\
(x +1)y &= B\\
(x + 1)(x + 8) &= B\\
x^2 + 9x + 8 &= B
\end{align}
Is what I get.
 
I would let $M$ be the age of Mark now, and we know Amy is 8. From the problem, we may then state:

$$9M=m^2$$

$$8(M+1)=4(2M+2)=n^2$$

Thus, we know $M$ must be a perfect square and $2M+2$ must also be a perfect square. Can you find such a number $M$?
 
From the statement above. it seems like will be having a Trial & Error method.
Is there another method you can share.

I actually have done it using trial & error which consume most of my time about 2 hours. I arise with the answer of : Mark = 49

what I want to know is real Algebra method, with equations.
hope you could help me. thank you.
 
My trial & Error Method
49 (8+1) = 441 where in \sqrt{441} = 21

(49+1) 8 = 400 where in \sqrt{400} = 20
 
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