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"mod 2" was missing in the REMARK.
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A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

for $\ s\equiv t\equiv 1,\ $$\ s\equiv t\equiv 1 \mod 2,\ $ and $\ \gcd(s\ t) = 1$.

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

for $\ s\equiv t\equiv 1,\ $ and $\ \gcd(s\ t) = 1$.

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

for $\ s\equiv t\equiv 1 \mod 2,\ $ and $\ \gcd(s\ t) = 1$.

just in case--the abc assumptions
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A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

for $\ s\equiv t\equiv 1,\ $ and $\ \gcd(s\ t) = 1$.

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

for $\ s\equiv t\equiv 1,\ $ and $\ \gcd(s\ t) = 1$.

Format (vertical space)
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A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

A warm-up:

$$ \left(\frac{s^n-t^n}2\right)^2 +\ (s\cdot t)^n\ =\ \left(\frac{s^n+t^n}2\right)^2 $$

provides integer solutions for $\,\ s\equiv t \mod 2 $.

(Sorry, I couldn't help it).

 

REMARK   More generally, going in the abc direction:

$$ \left(\frac{s^m-t^n}2\right)^2 +\ s^m\cdot t^n\ =\ \left(\frac{s^m+t^n}2\right)^2 $$

general exponent
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